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


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

Let $\frac{5 x}{(x+1)(x+2)(x-2)}=\frac{A}{(x+1)}+\frac{B}{(x+2)}+\frac{C}{(x-2)}$

$5 x=A(x+2)(x-2)+B(x+1)(x-2)+C(x+1)(x+2)$    ...(1)

Substituting = −1, −2, and 2 respectively in equation (1), we obtain

$A=\frac{5}{3}, B=-\frac{5}{2}$, and $C=\frac{5}{6}$

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

$\Rightarrow \int \frac{5 x}{(x+1)\left(x^{2}-4\right)} d x=\frac{5}{3} \int \frac{1}{(x+1)} d x-\frac{5}{2} \int \frac{1}{(x+2)} d x+\frac{5}{6} \int \frac{1}{(x-2)} d x$

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


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