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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,

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

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

#### 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

#### 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

# 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.

#### 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 $$A=1,B=0,C=2$$

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

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

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

#### 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

#### 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

#### 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

# $$\int\dfrac{Ax+B}{ax^2+bx+c}$$  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 $$t=2x-1$$)

#### 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 $$x=t^2-2$$

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

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

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

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

#### 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