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Integration Using Partial Fractions

Learn advanced methods of Integration by using partial fraction expansion & examples. Evaluate rational functions by using partial fraction & decomposition into partial fractions, finding the partial fractions coefficients. These partial fractions are easy to integrate.

Finding the Coefficients of Partial Fraction 

  • A rational function which is a ratio of two polynomial functions can be expressed as a sum of some simplex functions called its partial fractions. These partial fractions are easy to integrate.
  • Consider

\(\dfrac {1}{x+3} - \dfrac{2}{x+1}=\dfrac{(x+1)-2(x+3)}{(x+3)(x+1)}\)

\(\implies \)\(\dfrac {1}{x+3} - \dfrac{2}{x+1}=\dfrac{(-x-5)}{x^2+4x+3}\)

obtaining L.H.S. from R.H.S. is called finding partial fractions. Note that L.H.S. is easy to integrate.

  • \(\int \dfrac{-x-5}{x^2+4x+3}dx=\int\dfrac{1}{x+3}-\dfrac{2}{x+1}dx\).

\(\ell n |x+3|-2\ell n|x+1|+C\)

 

 

  • Consider a rational function

\(f(x)=\dfrac{P(x)}{Q(x)}\) where P and Q are polynomials and degree P < degree Q, such rational functions are called proper.

Case I

The denominator \(Q(x)\) is a product of distinct linear factors.

If \(Q(x)=(a_1x+b_1)(a_2x+b_2).....(a_nx+b_n)\) where no factor is repeated then

\(\dfrac{P(x)}{Q(x)}=\dfrac{A_1}{a_1x+b_1}+\dfrac{A_2}{a_2x+b_2}+......+\dfrac{A_n}{a_nx+b_n}\)

where, \(A_1,A_2,A_3....A_n\)  are constants and can be determined. These are called partial fractions coefficients.

Now to find \(A_1,A_2,A_3....A_n\), take L.C.M. in the R.H.S. and compare the numerator polynomials on both sides.

e.g.,

Consider,     \(\dfrac{7x-6}{(2x-1)(x-3)}\)

We decompose this into the partial fraction.

\(\dfrac{7x-6}{(2x-1)(x-3)}=\dfrac{A}{2x-1}+\dfrac{B}{x-3}\)

\(=\dfrac{A(x-3)+B(2x-1)}{(2x-1)(x-3)}\)

\(=\dfrac{(A+2B)x+(-3A-B)}{(2x-1)(x-3)}\)

Now, compare the numerator on both sides.

\(7x-6=x(A+2B)+(-3A-B)\)

\(\therefore A+2B=7\) ... (1)         (Coefficient of \(x\) should be equal on both sides)

and    \(-3A-B=-6\) ... (2)

solving (1) and (2) for A and B we get,

\(A=1\) , \(B=3\)

\(\therefore\) \(\dfrac{7x-6}{(2x-1)(x-3)}=\dfrac{1}{2x-1}+\dfrac{3}{x-3}\) .

\(\therefore\)  \(\displaystyle\int\dfrac{7x-6}{(2x-1)(x-3)}dx=\int\left(\dfrac{1}{2x-1}+\dfrac{3}{x-3}\right)dx\)

\(=\dfrac{\ell n(2x-1)}{2}+3\ell n(x-3)+C\)

Case 2

The denominator \(Q(x)\)  is a product of linear factors only, some or all of them may be repeated.

If \(Q(x)=(a_1x+b_1)^r\;(a_2x+b_2)^m\;......\)

then we assume the partial fraction as

\(\dfrac{P(x)}{Q(x)}=\dfrac{A_1}{(a_1x+b_1)}+\dfrac{A_2}{(a_1x+b_1)^2}+....\dfrac{A_r}{(a_1x+b_1)^r}+\dfrac{B_1}{(a_2x+b_2)}+\dfrac{B_2}{(a_2x+b_2)^2}+....\dfrac{B_m}{(a_2x+b_2)^m}\)

The factor which is repeated r times will have r terms in the partial fraction assumption.

Now to find \(A_1,A_2,A_3,.....,A_n\) take L.C.M. on the right-hand side and compare numerator polynomial on both sides.

e.g., Consider

\(\dfrac{1}{(x-1)^2(x+1)}=\dfrac{A}{x-1}+\dfrac{B}{(x-1)^2}+\dfrac{C}{x+1}\)

\(\dfrac{1}{(x-1)^2(x+1)}=\dfrac{A(x-1)(x+1)+B(x+1)+C(x-1)^2}{(x-1)^2(x+1)}\)

\(\Rightarrow\) \(1=A(x^2-1)+B(x+1)+C(x^2-2x+1)\)

\(\Rightarrow\) \(1=x^2(A+C)+x(B-2C)+(-A+B+C)\)

Compare coefficient of \(x^2,\;x\) and constant

\(A+C=0\)

\(B-2C=0\)

\(-A+B+C=1\)

Solving for \(A,B,C\) we get,

\(A=-\dfrac{1}{4},\;B=\dfrac{1}{2},\;C=\dfrac{1}{4}\)

\(\therefore\) \(\displaystyle\int\dfrac{1}{(x-1)^2(x+1)}dx=\int\left(\dfrac{-1}{4(x-1)}+\dfrac{1}{2(x-1)^2}+\dfrac{1}{4(x+1)}\right)dx\\ =-\dfrac{1}{4}\ell n|x-1|-\dfrac{1}{2(x-1)}+\dfrac{1}{4}\ell n|x+1|+C\)

 

Illustration Questions

Find  \(I=\displaystyle\int\dfrac{x}{(x+1)(2x+1)}dx\).

A \(I=\dfrac{-\ell n|2x+1|}{2}+\ell n |x+1|+C\)

B \(I={\ell n|3x+2|}-\ell n |x+1|+C\)

C \(I={-\ell n|5x+3|}-\ell n |x|+C\)

D \(I={\ell n|5x+4|}-\ell n |x+2|+C\)

×

Consider the perfect fraction decomposition of integral

\(\dfrac{x}{(2x+1)(x+1)}=\dfrac{A}{2x+1}+\dfrac{B}{x+1}\)

Compare numerators after taking L.C.M.

\(\dfrac{x}{(2x+1)(x+1)}=\dfrac{A(x+1)+B(2x+1)}{(2x+1)(x+1)}\)

\(\implies x=A(x+1)+B(2x+1)\)

\(\implies x=x(A+2B)+(A+B)\)

Compare the coefficients of \(x\) and constant on both sides.

\(\implies 1=A+2B \; \,\&\:\, 0 =A+B\)

Solving for \(A\) and \(B\) we get,

\(B=1\: \& \: A=-1\)

\(\therefore I = \int\dfrac{x}{(2x+1)(x+1)}dx=\int\dfrac{-1}{2x+1}+\int\dfrac{1}{x+1}dx\)

\(=\dfrac{-\ell n|2x+1|}{2}+\ell n|x+1|+C\)

Find  \(I=\displaystyle\int\dfrac{x}{(x+1)(2x+1)}dx\).

A

\(I=\dfrac{-\ell n|2x+1|}{2}+\ell n |x+1|+C\)

.

B

\(I={\ell n|3x+2|}-\ell n |x+1|+C\)

C

\(I={-\ell n|5x+3|}-\ell n |x|+C\)

D

\(I={\ell n|5x+4|}-\ell n |x+2|+C\)

Option A is Correct

Illustration Questions

Evaluate   \(I=\displaystyle\int\dfrac{2x^2+4x+3}{(x+2)(x+1)(2x+1)}dx\)

A \(I=\ell n |x+1|+3\ell n |x+2|-\dfrac{1}{2} \ell n (2x+1)+C\)

B \(I = \ell n|5x-1|-\ell n|x-2|+5|x+1|+C\)

C \(I = \ell n|2x^2+4x+3|-\ell n|x|+C\)

D \(I = \ell n|x+2|-\ell n|x+1|+\ell n |2x+1|+C\)

×

Consider the partial fraction decomposition of the integrand

\(\dfrac{2x^2+4x+3}{(x+2)(x+1)(2x+1)}=\dfrac{A}{x+2}+\dfrac{B}{x+1}+\dfrac{C}{2x+1}\)

Compare numerators after taking L.C.M. as both sides,

\(\dfrac{2x^2+4x+3}{(x+2)(x+1)(2x+1)}=\dfrac{A(x+1)(2x+1)+B(x+2)(2x+1)+C(x+2)(x+1)}{(x+2)(x+1)(2x+1)}\)

\(\implies 2x^2+4x+3=A(x+1)(2x+1)+B(x+2)(2x+1)+C(x+2)(x+1)\)

\(\implies 2x^2+4x+3=x^2(2A+2B+C)+x(3A+5B+3C)+A+2B+2C\)

Compare the coefficient of \(x^2,x\) and constraint on both side

\(\implies 2A+2B+C =2\)

\(3A+5B+3C=4\)

\(A+2B+2C=3\)

Solving for A, B & C we get,

A = 1, B = –1, C = 2

 

\(\therefore\) \(I=\displaystyle\int\dfrac{2x^2+4x+3}{(x+2)(x+1)(2x+1)}dx=\int\left(\dfrac{1}{x+2}-\dfrac{1}{x+1}+\dfrac{2}{2x+1}\right)dx\)

\(= \ell n|x+2|-\ell n|x+1|+\ell n |2x+1|+C\)

Evaluate   \(I=\displaystyle\int\dfrac{2x^2+4x+3}{(x+2)(x+1)(2x+1)}dx\)

A

\(I=\ell n |x+1|+3\ell n |x+2|-\dfrac{1}{2} \ell n (2x+1)+C\)

.

B

\(I = \ell n|5x-1|-\ell n|x-2|+5|x+1|+C\)

C

\(I = \ell n|2x^2+4x+3|-\ell n|x|+C\)

D

\(I = \ell n|x+2|-\ell n|x+1|+\ell n |2x+1|+C\)

Option D is Correct

Illustration Questions

Evaluate   \(I=\displaystyle\int\dfrac{(2x^2-7x+7)}{(x-2)^2(x-1)}dx\)

A \(I=\ell n|x-2|-\ell n|x-5|+C\)

B \(I=\dfrac{-1}{2x^2-7x+7}+\dfrac{1}{x-1}+\ell n|x-2|+C\)

C \(I=\ell n(x-2)^2+\ell n|x-1|+C\)  

D \(I=\dfrac{-1}{x-2}+2\ell n|x-1|+C\)

×

When \(Q(x)\) has repeated under factors \(\dfrac{P(x)}{Q(x)}\) will have partial fraction of the form in which factor, which is repeated r times, will have r terms.

In this case,

\(\dfrac{2x^2-7x+7}{(x-2)^2(x-1)}=\dfrac{A}{x-2}+\dfrac{B}{(x-2)^2}+\dfrac{C}{x-1}\)

Compare numerators on both sides after taking L.C.M.

\(\dfrac{2x^2-7x+7}{(x-2)^2(x-1)}=\dfrac{A(x-2)(x-1)+B(x-1)+C(x-2)^2}{(x-2)^2(x-1)}\)

\(\implies2x^2-7x+7=A(x-2)(x-1)+B(x-1)+C(x-2)^2\)

\(\implies2x^2-7x+7=A(x^2-3x+2)+B(x-1)+C(x^2-4x+4)\)

\(\implies 2x^2-7x+7=x^2(A+C)+x(-3A+B-4C)+(2A-B+4C)\)

Compare the coefficient of \(x^2,\;x\) and constant on both sides.

\(A+C=2\)

\(-3A+B-4C=-7\)

\(2A-B+4C=7\)

Solving for \(A,B,C\) we get

\(A=0,\;B=1,\;C=2\)

\(\displaystyle\therefore I=\int\dfrac{2x^2-7x+7}{(x-2)^2(x-1)}dx=\int\left(\dfrac{0}{x-2}+\dfrac{1}{(x-2)^2}+\dfrac{2}{x-1}\right)dx\)

\(=0+\dfrac{(x-2)^{-1}}{-1}+2\ell n|x-1|+C\)

\(=\dfrac{-1}{x-2}+2\ell n|x-1|+C\)

Evaluate   \(I=\displaystyle\int\dfrac{(2x^2-7x+7)}{(x-2)^2(x-1)}dx\)

A

\(I=\ell n|x-2|-\ell n|x-5|+C\)

.

B

\(I=\dfrac{-1}{2x^2-7x+7}+\dfrac{1}{x-1}+\ell n|x-2|+C\)

C

\(I=\ell n(x-2)^2+\ell n|x-1|+C\)
 

D

\(I=\dfrac{-1}{x-2}+2\ell n|x-1|+C\)

Option D is Correct

Illustration Questions

Evaluate  \(I=\displaystyle\int\dfrac{1}{x^3(x+1)}dx\)

A \(I=\ell n|x|+\dfrac{1}{x}-\dfrac{1}{2x^2}-\ell n|x+1|+C\)

B \(I=\ell n\;x^2-2\ell n|x+1|+\dfrac{1}{x}+C\)

C \(I=\dfrac{2}{x^2}-\dfrac{2}{x}+\ell n|x+2|+C\)

D \(I=\dfrac{1}{x^2}-\dfrac{2}{x^3}+\ell n|x|+C\)

×

When \(Q(x)\) has repeated under factors \(\dfrac{P(x)}{Q(x)}\) will have partial fraction of the form in which factor, which is repeated r times, will have r terms.

In this case,

\(\dfrac{1}{x^3(x+1)}=\dfrac{A}{x}+\dfrac{B}{x^2}+\dfrac{C}{x^3}+\dfrac{D}{x+1}\)

Compare the numerator on both sides after taking L.C.M. on R.H.S.

\(\dfrac{1}{x^3(x+1)}=\dfrac{A\;x^2(x+1)+B\;x(x+1)+C(x+1)+Dx^3}{x^3(x+1)}\)

\(\implies A(x^3+x^2)+B(x^2+x)+C(x+1)+Dx^3\)

\(\implies 1=x^3(A+D)+x^2(A+B)+x(B+C)+C\)

Compare the coefficient of \(x^3,\;x^2,\;x\) and constant on both sides.

\(A+D=0\)

\(A+B=0\)

\(B+C=0\)

\(C=1\)

Solving for \(A,B,C,D\) we get

\(A=1,\;B=-1,\;C=1,D=-1\)

\(\therefore I=\displaystyle\int\dfrac{1}{x^3(x+1)}dx=\int(\dfrac{1}{x}-\dfrac{1}{x^2}+\dfrac{1}{x^3}-\dfrac{1}{x+1})dx\)

\(=\ell n|x|+\dfrac{1}{x}-\dfrac{1}{2x^2}-\ell n|x+1|+C\)

 

Evaluate  \(I=\displaystyle\int\dfrac{1}{x^3(x+1)}dx\)

A

\(I=\ell n|x|+\dfrac{1}{x}-\dfrac{1}{2x^2}-\ell n|x+1|+C\)

.

B

\(I=\ell n\;x^2-2\ell n|x+1|+\dfrac{1}{x}+C\)

C

\(I=\dfrac{2}{x^2}-\dfrac{2}{x}+\ell n|x+2|+C\)

D

\(I=\dfrac{1}{x^2}-\dfrac{2}{x^3}+\ell n|x|+C\)

Option A is Correct

Integrals having Difference of Squares in Denominator

\(I=\displaystyle\int\dfrac{dx}{x^2-a^2}\)

Consider,

\(\dfrac{1}{x^2-a^2}=\dfrac{1}{(x-a)(x+a)}=\dfrac{A}{x-a}+\dfrac{B}{x+a}\)

\(\implies \dfrac{1}{(x-a)(x+a)}=\dfrac{A(x+a)+B(x-a)}{(x-a)(x+a)}\)

Compare the numerator \(\implies \) \(1=A(x+a)+B(x-a)\)

\(\implies\;1=x(A+B)+aA-aB\)

Compare coefficient of x and constant

\(\implies A+B=0,\,\,aA-aB=1\)

\(\implies A=\dfrac{1}{2a},\;B=\dfrac{-1}{2a}\)

\(\therefore \;I=\displaystyle\int\dfrac{1}{x^2-a^2}dx=\int\dfrac{1}{2a(x-a)}-\dfrac{1}{2a(x+a)}dx\)

\(=\dfrac{1}{2a}\ell n|x-a|-\dfrac{1}{2a}\ell n|x+a|+C\)

\(=\dfrac{1}{2a}\ell n\left|\dfrac{x-a}{x+a}\right|+C\)   (Use \(\ell n\dfrac{x}{y}=\ell n \;x-\ell n\;y\))

the above formula can be used directly in the integral.

 

Illustration Questions

Evaluate \(I=\displaystyle\int\dfrac{dx}{x^2-16}\)

A \(I=2sin^3x-cos^2x+x+c\)

B \(I=x^4-5x^3+\dfrac{1}{4}+C\)

C \(I=\dfrac{1}{2}\ell n\left|\dfrac{x-8}{x-5}\right|+C\)

D \(I=\dfrac{1}{8}\ell n\left|\dfrac{x-4}{x+4}\right|+C\)

×

\(\displaystyle\int\dfrac{1}{x^2-a^2}dx=\dfrac{1}{2a}\ell n\left|\dfrac{x-a}{x+a}\right|+C\)

\(\displaystyle\int\dfrac{1}{x^2-16}dx=\dfrac{1}{2×4}\ell n\left|\dfrac{x-4}{x+4}\right|+C\)       (a=4)

\(=\dfrac{1}{8}\ell n\left|\dfrac{x-4}{x+4}\right|+C\)

Evaluate \(I=\displaystyle\int\dfrac{dx}{x^2-16}\)

A

\(I=2sin^3x-cos^2x+x+c\)

.

B

\(I=x^4-5x^3+\dfrac{1}{4}+C\)

C

\(I=\dfrac{1}{2}\ell n\left|\dfrac{x-8}{x-5}\right|+C\)

D

\(I=\dfrac{1}{8}\ell n\left|\dfrac{x-4}{x+4}\right|+C\)

Option D is Correct

Partial Fractions having Irreducible Quadratic Factors in Denominator

\(Q(x)\) contains linear as well as irreducible quadratic factors which are distinct.

If \(Q(x)\) has factor \(ax^2+bx+c\) (where \(b^2-4ac<0\)) then the expression \(\dfrac{P(x)}{Q(x)}\) will have a factor of form \(\dfrac{Ax+B}{ax^2+bx+c}\).

where \(A,B\) are partial fraction coefficients.

Suppose we,  Evaluate \(I=\displaystyle\int\dfrac{x^2+2x-2}{(x-3)(x^2+4)}\)

Consider,

\(\dfrac{x^2+2x-2}{(x-3)(x^2+4)}=\dfrac{A}{x-3}+\dfrac{Bx+c}{x^2+4}\)

Compare the numerator on both sides after taking L.C.M. of R.H.S.

\(\implies x^2+2x-2=x^2(A+B)+x(C-3B)+4A-3C\)

Compare the coefficient of \(x^2,\;x\) and constant on both sides

\(\implies A+B=1,\;\;C-3B=2,\;\;4A-3C=-2\)

Solving for \(A,B,C\) we get \(A=1,B=0,C=2\)

\(\therefore\;I=\displaystyle\int\dfrac{x^2+2x-2}{(x-3)(x^2+4)}dx=\int\dfrac{1}{x-3}+\dfrac{2}{x^2+4}dx \)

\(=\ell n|x-3|+\dfrac{2}{2}tan^{-1}\dfrac{x}{2}+C\)

\(=\ell n|x-3|+tan^{-1}\dfrac{x}{2}+C\)

Illustration Questions

Evaluate \(I=\displaystyle\int\dfrac{x^2+2x+5}{(x-2)(x^2+9)}\)

A \(I=2sin\;x-x^2+x^3+C\)

B \(I= \ell n|x^2+2x+5|-\ell n|x|+C\)

C \(I=\ell n|x-2|+\dfrac{2}{3}tan^{-1}\dfrac{x}{3}+C\)

D \(I= \ell n|x^2+9|-\ell n|x|+C\)

×

\(\dfrac{P(x)}{Q(x)}=\dfrac{x^2+2x+5}{(x-2)(x^2+9)}=\dfrac{A}{x-2}+\dfrac{Bx+C}{x^2+9}\)

Compare the numerators after taking L.C.M. on both sides.

\(\dfrac{x^2+2x+5}{(x-2)(x^2+9)}=\dfrac{A(x^2+9)+(Bx+C)(x-2)}{(x-2)(x^2+9)}\)

\(\implies\;x^2+2x+5=A(x^2+9)+(Bx+C)(x-2)\)

\(\implies\;x^2+2x+5=x^2(A+B)+x(C-2B)+9A-2C\)

Compare the coefficient of \(x^2,\;x\) and constant on both sides

\(\implies A+B=1,\;\;C-2B=2,\;\;9A-2C=5\)

Solving for \(A,B,C\) we get

\(A=1,B=0,C=2\)

\(\therefore\;I=\displaystyle\int\dfrac{x^2+2x-5}{(x-2)(x^2+9)}dx=\int\left(\dfrac{1}{x-2}+\dfrac{2}{x^2+9}\right)dx \)

\(=\ell n|x-2|+\dfrac{2}{3}tan^{-1}\dfrac{x}{3}+C\)

Evaluate \(I=\displaystyle\int\dfrac{x^2+2x+5}{(x-2)(x^2+9)}\)

A

\(I=2sin\;x-x^2+x^3+C\)

.

B

\(I= \ell n|x^2+2x+5|-\ell n|x|+C\)

C

\(I=\ell n|x-2|+\dfrac{2}{3}tan^{-1}\dfrac{x}{3}+C\)

D

\(I= \ell n|x^2+9|-\ell n|x|+C\)

Option C is Correct

Evaluation of Integrals of the Form

\(\int\dfrac{Ax+B}{ax^2+bx+c}\)  where \(b^2-4ac<0\)

To evaluate the integral of the above form first make a perfect square in the denominator and then make a substitute which will make the integral of the form.

\(\int\dfrac{Cx+D}{x^2+a^2}dx=C\int\dfrac{x}{x^2+a^2}dx+D\int\dfrac{1}{x^2+a^2}dx\)

\(\int\dfrac{x}{x^2+a^2}dx\)  This is expressed in terms of the logarithm (Put \(x^2+a^2=t\)).

\(\int\dfrac{1}{x^2+a^2}dx\)  This is expressed in terms of \(tan^{-1}\).

e.g. Suppose we evaluate

\(I=\int\dfrac{x+2}{4x^2-4x+3}dx\)

\(4x^2-4x+3=(2x-1)^2+2\)     (Make perfect square)

Put \(2x-1=t\;\implies2dx=dt\)

Also  \(x=\dfrac{t+1}{2}\)

\(\therefore\;I=\int\dfrac{\dfrac{t+1}{2}+2}{t^2+2}\int\dfrac{dt}{2}=\dfrac{1}{4}\int\dfrac{t+5}{t^2+2}dt \)

\(=\dfrac{1}{4}\int\dfrac{t}{t^2+2}dt+=\dfrac{5}{4}\int\dfrac{1}{t^2+2}dt \)

\(=\dfrac{1}{2×4}\ell n|t^2+2|+\dfrac{5}{4×\sqrt{2}}tan^{-1}\dfrac{t}{\sqrt{2}}+C\)

\(\dfrac{1}{8}\ell n|4x^2-4x+3|+\dfrac{5}{{4}\sqrt{2}}tan^{-1}\left(\dfrac{2x-1}{\sqrt{2}}\right)+C \)  (put back \(t=2x-1\))

Illustration Questions

Evaluate \(I=\int\dfrac{x+3}{x^2+2x+5}dx\)

A \(I=\ell n|x+3|-2(x^2+2x+5)+C\)

B \(I=\dfrac{1}{2}\ell n|x^2+2x+5|+tan^{-1}\dfrac{x+1}{2}+C\)

C \(I=2cos^2x-\dfrac{1}{x}+tan\;x+C\)

D \(I=2\ell n|x-5|-\ell n|x-3|+C\)

×

\(I=\int\dfrac{x+3}{x^2+2x+5}dx\)

Complete the square for the expression  \(x^2+2x+5\)

\(x^2+2x+5=(x+1)^2+4\)

Put  \(x+1=t\) \(\implies dx=dt\)  and  \(x=t-1\)

\(\therefore I= \int\dfrac{(t-1)+3}{t^2+4}dt= \int\dfrac{t+2}{t^2+4}dt\)

\(= \int\dfrac{t}{t^2+4}dt+ \int\dfrac{2}{t^2+4}dt\)

\(= \dfrac{\ell n\;|t^2+4|}{2}+\dfrac{2}{2}tan^{-1}\dfrac {t}{2}+C\)

Put back  \(t=x+1\)

\(\implies I=\dfrac{1}{2}\ell n|x^2+2x+5|+tan^{-1}(\dfrac{x+1}{2})+C\)

Evaluate \(I=\int\dfrac{x+3}{x^2+2x+5}dx\)

A

\(I=\ell n|x+3|-2(x^2+2x+5)+C\)

.

B

\(I=\dfrac{1}{2}\ell n|x^2+2x+5|+tan^{-1}\dfrac{x+1}{2}+C\)

C

\(I=2cos^2x-\dfrac{1}{x}+tan\;x+C\)

D

\(I=2\ell n|x-5|-\ell n|x-3|+C\)

Option B is Correct

Integration of Partial Fractions using Rationalizing Substitution

  • Sometimes we come across integrands which contain irrational expressions,but by making suitable rationalising substitution we can convert this integral to rational functions.

e.g.,

Suppose

\(I=\displaystyle\int\dfrac{\sqrt{x+2}}{x}dx\)

We put  \(x+2=t^2\implies\;dx=2t\;dt\) 

and \(x=t^2-2\)

\(\therefore I=\displaystyle\int\dfrac{t×2t\;dt}{t^2-2}=2\int\dfrac{t^2}{t^2-2}dt\)

\(=2\displaystyle\int\dfrac{t^2-2+2}{t^2-2}dt=2\int\left(1+\dfrac{2}{t^2-2}\right)dt\)

\(=2\left[ t+\dfrac{2}{2\sqrt{2}}\ell n\dfrac{t-\sqrt{2}}{t+\sqrt{2}} \right]+C\)

\(=2\left[ \sqrt{x+2}+\dfrac{1}{\sqrt{2}}\ell n \left(\dfrac{\sqrt{x+2}-\sqrt{2}}{\sqrt{x+2}+\sqrt{2}} \right)\right]+C\)

Illustration Questions

Evaluate  \(I=\displaystyle\int\dfrac{dx}{x\sqrt{x+1}}\)

A \(I=\ell n \left|\dfrac{\sqrt{x+1}-1}{\sqrt{x+1}+1} \right|+c\)

B \(I=\ell n |x+1|-sin^2x+c\)

C \(I=\ell n(\sqrt{x+1})-2\ell n(\sqrt{x-1})+c\)

D \(I=\dfrac{1}{x^2}-\dfrac{1}{x^3}+c\)

×

\(I=\displaystyle\int\dfrac{dx}{x\sqrt{x+1}}\)

To rationalise the integrand we make the substitution

\(x+1=t^2\;\implies\;dx=2t\;dt\)

and \(x=t^2-1\)

\(\therefore I=\displaystyle\int\dfrac{2t\;dt}{(t^2-1)×t}=2\int\dfrac{dt}{t^2-1}\)

\(=2×\dfrac{1}{2}\ell n \left| \dfrac{t-1}{t+1}\right|+C\)

\(=\ell n \left| \dfrac{t-1}{t+1}\right|+C\)

Put back  \(t=\sqrt{x+1}\)

\(\Rightarrow I=\ell n \left|\dfrac{\sqrt{x+1}-1}{\sqrt{x+1}+1} \right|+c\)

Evaluate  \(I=\displaystyle\int\dfrac{dx}{x\sqrt{x+1}}\)

A

\(I=\ell n \left|\dfrac{\sqrt{x+1}-1}{\sqrt{x+1}+1} \right|+c\)

.

B

\(I=\ell n |x+1|-sin^2x+c\)

C

\(I=\ell n(\sqrt{x+1})-2\ell n(\sqrt{x-1})+c\)

D

\(I=\dfrac{1}{x^2}-\dfrac{1}{x^3}+c\)

Option A is Correct

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