Browse Questions

# Evaluate the following:$\int\frac{x^2dx}{(x^2+a^2)(x^2+b^2)}$

$\begin{array}{1 1} (A)\;\large\frac{1}{a^2-b^2}\bigg[ a \;\tan ^{-1}(\large\frac{x}{a})+b\;\tan ^{-1} \large(\frac{x}{b})\bigg]+c \\(B)\;\large\frac{1}{a^2+b^2}\bigg[ a \;\tan ^{-1}(\large\frac{x}{a})- b\;\tan ^{-1} \large(\frac{x}{b})\bigg]+c \\ (C)\;\large\frac{1}{a^2-b^2}\bigg[ a \;\tan ^{-1}(\large\frac{x}{a})- b\;\tan ^{-1} \large(\frac{x}{b})\bigg]+c \\ (D)\;\large\frac{1}{a+b}\bigg[ a \;\tan ^{-1}(\large\frac{x}{a})- b\;\tan ^{-1} \large(\frac{x}{b})\bigg]+c \end{array}$

Toolbox:
• A rational expression of the form $\large\frac{x}{(x+a)(x+b)}$ can be resolved in to partial fractions as $\large\frac{A}{x+a}+\frac{B}{x+b}$
• $\int \large\frac{dx}{x^2+a^2}=\frac{1}{a} $$\tan ^{-1} \bigg(\large\frac{x}{a}\bigg)+c Let I=\large\int\frac{x^2dx}{(x^2+a^2)(x^2+b^2)} \large\frac{x^2}{(x^2+a^2)(x^2+b^2)} can be resolved into partial fractions as \large\frac{x^2}{(x^2+a^2)(x^2+b^2)}=\frac{Ax+B}{x^2+a^2}+\frac{Cx+D}{x^2+b^2} Now let us evaluate the values for A,B,C and D x^2=(Ax+B)(x^2+b^2)+(Cx+D)(x^2+a^2) Now equating the coefficient of like terms first equating for x^3 0=A+C-------(1) =>A =-C Next equating for x^2 1=B+D-----(2) =>B=1-D Next equating for x 0=Ab^2+Ca^2-----(3) Next equating for constant term 0=Bb^2+Da^2-----(4) Substituting for A in equ (3) -Cb^2+Ca^2=0 C(a^2-b^2)=0=>C=0 Therefore A=0 Substituting for B in equ (4) (1-0)b^2+Da^2=0 b^2-Db^2+Da^2=0 D(a^2-b^2)=-b^2 =>\large\frac{-b^2}{a^2-b^2} Now 1-D=B substituting for D we get 1+\large\frac{b^2}{a^2-b^2}=B Therefore B=\large\frac{a^2-b^2+b^2}{a^2-b^2}=\frac{a^2}{a^2-b^2} Therefore \large\frac{x^2}{(x^2+a^2)(x^2+b^2)}=\large\frac{a^2}{(a^2-b^2)(x^2+a^2)}-\large\frac{b^2}{(a^2-b^2)(x^2+b^2)} Hence I=\large \frac{a^2}{a^2-b^2} \int \frac{dx}{x^2+a^2}-\frac{b^2}{a^2-b^2}\int \frac{dx}{x^2+b^2} This is of the form \int \large\frac{dx}{x^2+a^2}=\frac{1}{a}$$\tan ^{-1}\large(\frac{x}{a})+c$
On integration we get
$I=\large\frac{a^2}{a^2-b^2}.\frac{1}{a} $$\tan ^{-1}(\large\frac{x}{a})-\frac{b^2}{a^2-b^2} \times \frac{1}{b}$$\tan ^{-1} \large(\frac{x}{b})+c$
$=\large\frac{a}{a^2-b^2}$$\tan ^{-1}(\large\frac{x}{a})-\frac{b}{a^2-b^2}$$\tan ^{-1} \large(\frac{x}{b})+c$
$=\large\frac{1}{a^2-b^2}\bigg[ $$a \;\tan ^{-1}(\large\frac{x}{a})-$$b\;\tan ^{-1} \large(\frac{x}{b})\bigg]+c$