If int frac{1}{xleft[(log x)^2+4 log x-1right | vedclass

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

Question

(A) Let $I = \int \frac{1}{x[(\log x)^2 + 4 \log x - 1]} dx$. Substitute $\log x = t$,so $\frac{1}{x} dx = dt$. The integral becomes $I = \int \frac{1

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $I = \int \frac{1}{x[(\log x)^2 + 4 \log x - 1]} dx$. Substitute $\log x = t$,so $\frac{1}{x} dx = dt$. The integral becomes $I = \int \frac{1}{t^2 + 4t - 1} dt$. Complete the square in the denominator: $t^2 + 4t - 1 = (t+2)^2 - 5 = (t+2)^2 - (\sqrt{5})^2$. Using the standard integral formula $\int \frac{1}{u^2 - a^2} du = \frac{1}{2a} \log \left| \frac{u-a}{u+a} \right| + K$,we get: $I = \frac{1}{2\sqrt{5}} \log \left| \frac{(t+2) - \sqrt{5}}{(t+2) + \sqrt{5}} \right| + K$. Substituting $t = \log x$ back: $I = \frac{1}{2\sqrt{5}} \log \left| \frac{\log x + 2 - \sqrt{5}}{\log x + 2 + \sqrt{5}} \right| + K$. Comparing this with the given form,we find $A = \frac{1}{2\sqrt{5}}$,$B = 2 - \sqrt{5}$,and $C = 2 + \sqrt{5}$.

If $\int \frac{1}{x\left[(\log x)^2+4 \log x-1\right]} d x=A \log \left[\frac{\log x+B}{\log x+C}\right]+K$ where $K$ is the constant of integration,then

Similar Questions

The value of $\int \frac{1+x^{4}}{1+x^{6}} dx$ is

$\int \frac{1}{\cos x \sqrt{\cos 2 x}} \, dx = $ (where $C$ is a constant of integration.)

$\int \frac{x^{\frac{1}{3}}}{(1 + x^{\frac{2}{3}})^3} dx$ is equal to (where $C$ is the constant of integration).

If $\int \sqrt{x-\frac{1}{x}}\left(\frac{x^{2}+1}{x^{2}}\right) d x=\frac{2}{3}\left(x-\frac{1}{x}\right)^{k}+c$,then the value of $k$ is

The integral $\int \frac{\sin ^2 x \cos ^2 x}{\left(\sin ^5 x+\cos ^3 x \sin ^2 x+\sin ^3 x \cos ^2 x+\cos ^5 x\right)^2} \,d x$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module