JEE MAIN - Mathematics (2023 - 25th January Evening Shift - No. 11)
The integral $$16\int\limits_1^2 {{{dx} \over {{x^3}{{\left( {{x^2} + 2} \right)}^2}}}} $$ is equal to
$${{11} \over {12}} + {\log _e}4$$
$${{11} \over 6} + {\log _e}4$$
$${{11} \over {12}} - {\log _e}4$$
$${{11} \over 6} - {\log _e}4$$
Explanation
$I=\int \frac{d x}{x^{3}\left(x^{2}+2\right)^{2}}$
$=\frac{1}{4} \int \frac{x}{x^{2}+2} d x+\frac{1}{4} \int \frac{x}{\left(x^{2}+2\right)^{2}}-\frac{1}{4} \int \frac{d x}{x}+\frac{1}{4} \int \frac{d x}{x^{3}}$
$=\frac{1}{8} \ln \left(x^{2}+2\right)-\frac{\ln x}{4}-\frac{1}{8\left(x^{2}+2\right)}-\frac{1}{8 x^{3}}$
Now, $16 \int_{1}^{2} \frac{d x}{x^{3}\left(x^{2}+2\right)^{2}}=2 \ln 6-2 \ln 3-4 \ln 2+\frac{11}{6}$
$=\frac{11}{6}-\ln 4$
$=\frac{1}{4} \int \frac{x}{x^{2}+2} d x+\frac{1}{4} \int \frac{x}{\left(x^{2}+2\right)^{2}}-\frac{1}{4} \int \frac{d x}{x}+\frac{1}{4} \int \frac{d x}{x^{3}}$
$=\frac{1}{8} \ln \left(x^{2}+2\right)-\frac{\ln x}{4}-\frac{1}{8\left(x^{2}+2\right)}-\frac{1}{8 x^{3}}$
Now, $16 \int_{1}^{2} \frac{d x}{x^{3}\left(x^{2}+2\right)^{2}}=2 \ln 6-2 \ln 3-4 \ln 2+\frac{11}{6}$
$=\frac{11}{6}-\ln 4$
Comments (0)
