int_2^4 frac{log x^2}{log x^2+log (36-12x+x^2 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $I = \int_2^4 \frac{\log x^2}{\log x^2 + \log (36-12x+x^2)} dx$. Note that $36-12x+x^2 = (6-x)^2$. So,$I = \int_2^4 \frac{\log x^2}{\log x^2 +

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) Let $I = \int_2^4 \frac{\log x^2}{\log x^2 + \log (36-12x+x^2)} dx$. Note that $36-12x+x^2 = (6-x)^2$. So,$I = \int_2^4 \frac{\log x^2}{\log x^2 + \log (6-x)^2} dx$. Using the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$,we have $a+b = 2+4 = 6$. Replacing $x$ with $6-x$: $I = \int_2^4 \frac{\log (6-x)^2}{\log (6-x)^2 + \log (6-(6-x))^2} dx = \int_2^4 \frac{\log (6-x)^2}{\log (6-x)^2 + \log x^2} dx$. Adding the two expressions for $I$: $2I = \int_2^4 \frac{\log x^2 + \log (6-x)^2}{\log x^2 + \log (6-x)^2} dx = \int_2^4 1 dx$. $2I = [x]_2^4 = 4-2 = 2$. Therefore,$I = 1$.

$\int_2^4 \frac{\log x^2}{\log x^2+\log (36-12x+x^2)} dx$ is equal to

Similar Questions

$\int_0^1 f(1 - x) \, dx$ has the same value as which of the following integrals?

$\int_0^\pi \frac{\theta \sin \theta}{1+\cos ^2 \theta} d \theta$ is equal to

If $S_n = \int_0^{\frac{\pi}{2}} \frac{\sin((2n-1)x)}{\sin x} dx$ and $n$ is an integer,then $S_{n+1} - S_n =$

If $h(a) = h(b)$,the value of the integral $\int_a^b {[f(g(h(x)))]^{-1} f'(g(h(x))) \cdot g'(h(x)) \cdot h'(x) \, dx} = $

The value of the definite integral $\int_{\pi / 24}^{5 \pi / 24} \frac{d x}{1+\sqrt[3]{\tan 2 x}}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module