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  • explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 25

explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 25

Answers (1)

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Answer:

            \frac{-1}{14} \tan ^{-1} \frac{x}{2}+\frac{8}{35} \tan ^{-1} \frac{x}{5}+C

Hint:

            To solve this integration, we use partial fraction method   

Given:

            \int \frac{x^{2}+1}{\left(x^{2}+4\right)\left(x^{2}+25\right)} d x \\

Explanation:

Let

\begin{aligned} &I=\int \frac{x^{2}+1}{\left(x^{2}+4\right)\left(x^{2}+25\right)} d x \\ &\frac{x^{2}+1}{\left(x^{2}+4\right)\left(x^{2}+25\right)}=\frac{A x+B}{x^{2}+4}+\frac{C x+D}{x^{2}+25} \\ &x^{2}+1=(A x+B)\left(x^{2}+25\right)+(C x+D)\left(x^{2}+4\right) \\ &x^{2}+1=x^{3}(A+C)+x^{2}(B+D)+x(25 A+4 C)+(25 B+4 D) \end{aligned}

Comparing the coefficient

Coefficient of x^{3}

A+C=0                    (2)

A=-C                       (3)

Coefficient of x^{2}

B+D=1                    (4)

Coefficient of x

25A+4C=0

-21C+4C=0                               [From the equation (3)]

-21C=0

C=0                                   (5)

A=0                                   (6)

Constant term

25 B+4 D=1                   (7)

 Multiply the equation (4) by 4 and then subtract it from equation (7)

\begin{aligned} &25 B+4 D=1- \\ &4 B+4 D=4 \\ &\overline{21 B=-3} \\ &B=\frac{-1}{7} \end{aligned}

Equation (4)

\begin{aligned} &\frac{-1}{7}+D=1 \\ &D=1+\frac{1}{7} \\ &D=\frac{8}{7} \\ &\frac{\left(x^{2}+1\right)}{\left(x^{2}+4\right)\left(x^{2}+25\right)}=\frac{-1}{7\left(x^{2}+4\right)}+\frac{8}{7\left(x^{2}+25\right)} \end{aligned}

Thus

\begin{aligned} &I=\frac{-1}{7} \int \frac{1}{x^{2}+4} d x+\frac{8}{7} \int \frac{1}{x^{2}+25} d x \\ &I=\frac{-1}{7} \cdot\left(\frac{1}{2}\right) \tan ^{-1} \frac{x}{2}+\frac{8}{7} \cdot\left(\frac{1}{5}\right) \tan ^{-1}\left(\frac{x}{5}\right)+C \\ &I=\frac{-1}{14} \tan ^{-1} \frac{x}{2}+\frac{8}{35} \tan ^{-1} \frac{x}{5}+C \end{aligned}

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