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  • explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 41

explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 41

Answers (1)

Answer:

                \frac{1}{3} \tan ^{-1} x-\frac{1}{6} \tan ^{-1} \frac{x}{2}+C

Hint:

            To solve this integration, we use partial fraction method   

Given:

            \int \frac{1}{\left(x^{2}+1\right)\left(x^{2}+4\right)} d x \\

Explanation:

Let

\begin{aligned} &I=\int \frac{1}{\left(x^{2}+1\right)\left(x^{2}+4\right)} d x \\ &\frac{1}{\left(x^{2}+1\right)\left(x^{2}+4\right)}=\frac{A x+B}{x^{2}+1}+\frac{C x+D}{x^{2}+4} \\ &1=(A x+B)\left(x^{2}+4\right)+(C x+D)\left(x^{2}+1\right) \\ &1=x^{3}(A+C)+x^{2}(B+D)+(4 A+C) x+4 B+D \end{aligned}

Equating the similar terms

\begin{aligned} &A+C=0 \\ &A=-C \end{aligned}                       (1)                   

\begin{aligned} &B+D=0 \\ &B=-D \end{aligned}                       (2)

\begin{aligned} &4 A+C=0 \\ &-4 C+C=0 \end{aligned}                [From the equation (1)]

\begin{aligned} &-3 C=0 \\ &C=0 \\ &A=0 \end{aligned}                          [From the equation (1)]

\begin{aligned} &4 B+D=1 \\ &-4 D+D=1 \\ &-3 D=1 \end{aligned}

D=\frac{-1}{3}                               (3)

B=\frac{1}{3}                                  [From the equation (2)]

\begin{aligned} &\frac{1}{\left(x^{2}+1\right)\left(x^{2}+4\right)}=\frac{1}{3\left(x^{2}+1\right)}+\frac{(-1)}{3\left(x^{2}+4\right)} \\ &I=\frac{1}{3} \int \frac{1}{x^{2}+1} d x-\frac{1}{3} \int \frac{1}{x^{2}+4} d x \\ &I=\frac{1}{3} \tan ^{-1} x-\frac{1}{3} \cdot\left(\frac{1}{2}\right) \tan ^{-1} \frac{x}{2}+C \\ &I=\frac{1}{3} \tan ^{-1} x-\frac{1}{6} \tan ^{-1} \frac{x}{2}+C \end{aligned}

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