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  • explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 7

explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 7

Answers (1)

Answer:

            \frac{5}{3} \log |x+1|+\frac{5}{6} \log |x-2|-\frac{5}{2} \log |x+2|+C

Hint:

            To solve this integration, we use partial fraction method   

Given:

            \int \frac{5 x}{(x+1)\left(x^{2}-4\right)} d x

Explanation:

Let

\begin{aligned} &I=\int \frac{5 x}{(x+1)\left(x^{2}-4\right)} d x \\ &\left.I=\int \frac{5 x}{(x+1)(x-2)(x+2)} d x \ldots \text { (applying the } \text { formula } a^{2}-b^{2}\right) \end{aligned}

\begin{aligned} &\frac{5 x}{(x+1)(x-2)(x+2)}=\frac{A}{x+1}+\frac{B}{x-2}+\frac{C}{x+2} \\\\ &\frac{5 x}{(x+1)(x-2)(x+2)}=\frac{A(x-2)(x+2)+B(x+1)(x+2)+C(x+1)(x-2)}{(x+1)(x-2)(x+2)} \\\\ &5 x=A(x-2)(x+2)+B(x+1)(x+2)+C(x+1)(x-2) \end{aligned}

\begin{aligned} &\text { At } x=2 \\ &5(2)=0+B(3)(4)+0 \\ &10=12 B \\ &B=\frac{10}{12} \\ &B=\frac{5}{6} \end{aligned}

\begin{aligned} &\text { At } x=-2 \\ &5(-2)=0+0+C(-1)(-4) \\ &-10=4 C \\ &C=\frac{-10}{4} \\ &\text { At } x=-1 \end{aligned}

\begin{aligned} &5(-1)=A(-3)(1)+0+0 \\ &-5=-3 A \\ &\begin{array}{l} A=\frac{5}{3} \\\\ \end{array} \end{aligned}

\frac{5 x}{(x+1)(x-2)(x+2)}=\frac{5}{3(x+1)}+\frac{5}{6(x-2)}-\frac{5}{2(x+2)}

\begin{aligned} &I=\int\left(\frac{5}{3(x+1)}+\frac{5}{6(x-2)}-\frac{5}{2(x+2)}\right] d x \\ &I=\frac{5}{3} \int \frac{1}{x+1} d x+\frac{5}{6} \int \frac{1}{x-2} d x-\frac{5}{2} \int \frac{1}{x+2} d x \\ &I=\frac{5}{3} \log |x+1|+\frac{5}{6} \log |x-2|-\frac{5}{2} \log |x+2|+C \end{aligned}

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