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#### explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 50

$-\frac{1}{4 x}+\frac{7}{8} \tan ^{-1}\left(\frac{x}{2}\right)+C$

Hint:

To solve this integration, we use partial fraction method

Given:

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

Explanation:

Let

$I= \int \frac{2 x^{2}+1}{x^{2}\left(x^{2}+4\right)}dx$

Now let’s separate the fraction $\frac{2 x^{2}+1}{x^{2}\left(x^{2}+4\right)}$ through the partial fraction

Put

\begin{aligned} &4 A=1 \\ &A=\frac{1}{4} \\ &A+B=2 \\ &B=2-A \\ &B=2-A \\ &B=2-\frac{1}{4} \\ &B=\frac{7}{4} \end{aligned}

Equating both side

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

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