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.

- 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,

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

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

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

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

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

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.

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

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

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

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

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

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

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

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

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 \(t=2x-1\))

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

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

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

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

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

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

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