int frac{(log x-1)^2}{left[1+(log x)^2right]^ | vedclass

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

Question

(C) Let $I = \int \frac{(\log x-1)^2}{\left\{1+(\log x)^2\right\}^2} dx$. Substitute $\log x = t$,so $x = e^t$ and $dx = e^t dt$. Then $I = \int \frac

Vedclass · Accepted Answer

$\frac{x}{1+(\log x)^2}+C$

Vedclass · Answer

(C) Let $I = \int \frac{(\log x-1)^2}{\left\{1+(\log x)^2\right\}^2} dx$. Substitute $\log x = t$,so $x = e^t$ and $dx = e^t dt$. Then $I = \int \frac{(t-1)^2}{(1+t^2)^2} e^t dt = \int e^t \left( \frac{t^2-2t+1}{(1+t^2)^2} \right) dt$. We can rewrite the integrand as $\int e^t \left( \frac{t^2+1-2t}{(1+t^2)^2} \right) dt = \int e^t \left( \frac{1}{1+t^2} - \frac{2t}{(1+t^2)^2} \right) dt$. Using the formula $\int e^t (f(t) + f'(t)) dt = e^t f(t) + C$,where $f(t) = \frac{1}{1+t^2}$ and $f'(t) = -\frac{2t}{(1+t^2)^2}$. Thus,$I = e^t \left( \frac{1}{1+t^2} \right) + C$. Substituting back $t = \log x$,we get $I = \frac{e^{\log x}}{1+(\log x)^2} + C = \frac{x}{1+(\log x)^2} + C$.

$\int \frac{(\log x-1)^2}{\left[1+(\log x)^2\right]^2} d x=$ (where $C$ is constant of integration.)

Similar Questions

If $I = \int e^{\sin \theta} (\log \sin \theta + \operatorname{cosec}^2 \theta) \cos \theta \, d\theta$,then $I$ is equal to

$\int \left[ \frac{1}{\log x} - \frac{1}{(\log x)^2} \right] dx =$

If an antiderivative of $f(x)$ is $e^x$ and that of $g(x)$ is $\cos x,$ then $\int f(x) \cos x \, dx + \int g(x) e^x \, dx = $

If $\int e^x(x^3+x^2-x+4) dx = e^x f(x) + c$, then $f(1) =$

Assertion $(A)$: $\int_2^e \left(\frac{1}{\log_e x} - \frac{1}{(\log_e x)^2}\right) dx = e - 2 \log_2 e$
Reason $(R)$: $\int_a^b e^x (f(x) + f'(x)) dx = e^b f(b) - e^a f(a)$

Vedclass Test Series

Exam Paper Generator

Online Exam Module