int_1^3 frac{log x^2}{log left(16 x^2-8 x^3+x | vedclass

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

Question

(A) Let $I = \int_1^3 \frac{\log x^2}{\log \left(16 x^2-8 x^3+x^4\right)} dx$. Note that $16x^2 - 8x^3 + x^4 = x^2(16 - 8x + x^2) = x^2(4-x)^2$. Thus,

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) Let $I = \int_1^3 \frac{\log x^2}{\log \left(16 x^2-8 x^3+x^4\right)} dx$. Note that $16x^2 - 8x^3 + x^4 = x^2(16 - 8x + x^2) = x^2(4-x)^2$. Thus,the denominator is $\log(x^2(4-x)^2) = \log x^2 + \log(4-x)^2$. So,$I = \int_1^3 \frac{\log x^2}{\log x^2 + \log(4-x)^2} dx$. Using the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$,we have $a+b = 1+3 = 4$. Replacing $x$ with $(4-x)$,we get $I = \int_1^3 \frac{\log(4-x)^2}{\log(4-x)^2 + \log x^2} dx$. Adding the two expressions for $I$: $2I = \int_1^3 \frac{\log x^2 + \log(4-x)^2}{\log x^2 + \log(4-x)^2} dx = \int_1^3 1 dx = [x]_1^3 = 3-1 = 2$. Therefore,$I = 1$.

$\int_1^3 \frac{\log x^2}{\log \left(16 x^2-8 x^3+x^4\right)} d x=\ldots$

Similar Questions

If $\int_0^{2024 \pi} \frac{2023^{\sin ^2 x}}{2023^{\sin ^2 x}+2023^{\cos ^2 x}} d x=k$,then $\left(\frac{2 k}{\pi}+1\right)=$

The value of $\int_{-\pi}^{\pi} \frac{\cos^2 x}{1 + a^x} dx, a > 0$ is

$\int_0^{400 \pi} \sqrt{1-\cos 2 x} \, dx =$ (in $\sqrt{2}$)

$\int_0^{\infty} (x^{12} + x^{-12}) \frac{\log x}{x} dx =$

For any integer $n$,the value of the integral $\int_0^\pi e^{\cos^2 x} \cos^3(2n+1)x \, dx$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module