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  • explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 58

explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 58

Answers (1)

Answer:

                \frac{a}{a^{2}-b^{2}} \tan ^{-1} \frac{x}{a}+\frac{b}{b^{2}-a^{2}} \tan ^{-1} \frac{x}{b}+C

Hint:

            To solve this integration, we use partial fraction method   

Given:

            \int \frac{x^{2}}{\left(x^{2}+a^{2}\right)\left(x^{2}+b^{2}\right)} d x

Explanation:

Let

I=\int \frac{x^{2}}{\left(x^{2}+a^{2}\right)\left(x^{2}+b^{2}\right)} d x

\frac{x^{2}}{\left(x^{2}+a^{2}\right)\left(x^{2}+b^{2}\right)}=\frac{A x+B}{\left(x^{2}+a^{2}\right)}+\frac{C x+D}{\left(x^{2}+b^{2}\right)} \\        (i)

x^{2}=(A x+B)\left(x^{2}+b^{2}\right)+(C x+D)\left(x^{2}+a^{2}\right)                                    

On comparing coefficient

x^{2}Coefficient             

1=B+D         (1)

x Coefficient              

0=A b^{2}+C a^{2}               (2)

Constant

0=B b^{2}+D a^{2}              (3)

\begin{aligned} &x^{3} \text { Coefficient }\\ &0=A+C \end{aligned}                       (4)

Solving (2) and (4)

\begin{aligned} &A b^{2}+C a^{2}=0 \\ &A+C=0 \\ &C=-A \\ &A b^{2}-A a^{2}=0 \\ &A\left(b^{2}-a^{2}\right)=0 \\ &A=0, C=0 \end{aligned}

B+D=1                    (1)

B b^{2}+D a^{2}=0              (3)

Multiply (1) by b^{2}and subtract (3)

\begin{aligned} &B b^{2}+D b^{2}=b^{2}- \\ &B b^{2}+D a^{2}=0 \\ &\overline{D\left(b^{2}-a^{2}\right)=b^{2}} \end{aligned}

\begin{aligned} &D=\frac{b^{2}}{b^{2}-a^{2}} \\ &B=1-D \\ &B=1-\frac{b^{2}}{b^{2}-a^{2}} \\ &B=\frac{-a^{2}}{b^{2}-a^{2}} \end{aligned}

(i) Becomes
\begin{aligned} &\frac{x^{2}}{\left(x^{2}+a^{2}\right)\left(x^{2}+b^{2}\right)} d x=\frac{-a^{2}}{b^{2}-a^{2}} \int \frac{1}{x^{2}+a^{2}} d x+\frac{b^{2}}{b^{2}-a^{2}} \int \frac{1}{x^{2}+b^{2}} d x \\ &=\frac{a^{2}}{a^{2}-b^{2}} \cdot \frac{1}{a} \tan ^{-1} \frac{x}{a}+\frac{b^{2}}{b^{2}-a^{2}} \cdot \frac{1}{b} \tan ^{-1} \frac{x}{b} \\ &=\frac{a}{a^{2}-b^{2}} \tan ^{-1} \frac{x}{a}+\frac{b}{b^{2}-a^{2}} \tan ^{-1} \frac{x}{b}+C \end{aligned}

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