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

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

Answers (1)

Answer:

                -\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}

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