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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)

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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