The integral ∫24(logx^2)/(logx^2+log6-x^2)dx is equal to

Correct answer: (D) 1. I=∫24(logx^2)/(logx^2+log6-x^2)dx I=∫24(log6-x^2)/(log6-x^2+logx^2)dx ∵∫abfxdx=∫abfa+b-xdx 2I=∫24(logx^2+log6-x^2)/(log6-x^2+logx^2)dx 2I=∫241dx 2I=x24 ∴ I=1

Q54 / 90 · JEE Main 2015 (04 Apr) · Mathematics

The integral $\int_{2}^{4} \frac{log x^{2}}{log x^{2} + log {(6 - x)}^{2}} d x$ is equal to

A
$6$
B
$2$
C
$4$
Correct answer
D
$1$

Solution

$I = \int_{2}^{4} \frac{log x^{2}}{log x^{2} + log {(6 - x)}^{2}} d x$

$I = \int_{2}^{4} \frac{log {(6 - x)}^{2}}{log {(6 - x)}^{2} + log x^{2}} d x$

$[∵ \int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x]$

$2 I = \int_{2}^{4} \frac{log x^{2} + log {(6 - x)}^{2}}{log {(6 - x)}^{2} + log x^{2}} d x$

$2 I = \int_{2}^{4} 1 dx$

$2 I = {[x]}_{2}^{4}$

$∴ I = 1$

Hence, the correct answer is option: 'D'.

The integral ∫24logx2logx2+log6-x2dx is equal to

Solution

Relevant Questions

The integral $\int_{2}^{4} \frac{log x^{2}}{log x^{2} + log {(6 - x)}^{2}} d x$ is equal to