∫(1+(logx)2)2(logx−1)2dx=∫[(1+(logx)2)1−(1+(logx)2)22logx]dx=∫[1+t2et−(1+t2)22tet]dt put logx=t⇒dx=etdt∫et[1+t21−(1+t2)22t]dt=1+t2et+c=1+(logx)2x+c
∫ (log x-1) (1+(log x)^2 . ^2 d x is equal to
JEE Main 2005 — Mathematics Calculus
∫{(1+(logx)2(logx−1)}2dx is equal to
Held on 30 Apr 2005 · Verified 6 Jul 2026.
Options
- A
(logx)2+1logx+C
- B
x2+1x+C
- C
1+x2xex+C
- D
(logx)2+1x+C
Did you get this right?
Sign in to track your attempts and accuracy.
Your note
Sign in to keep a private note on this question. Nothing you write is ever public.
JEE Main Calculus in other years
More JEE Main Calculus questions
If $\int\left(\frac{1-5 \cos ^{2} x}{\sin ^{5} x \cos ^{2} x}\right) d x=f(x)+\mathrm{C}$, where C is the constant of integration, then $f\left(\frac{\pi}{6}\right)-f\left(\frac{\pi}{4}\right)$ is equal to
If the solution curve $y=f(x)$ of the differential equation $\left(x^{2}-4\right) y^{\prime}-2 x y+2 x\left(4-x^{2}\right)^{2}=0, x>2$, passes through the point $(3,15)$, then the local maximum value of $f$ is $\_\_\_\_$.
Let $[\cdot]$ be the greatest integer function. If $\alpha=\int_{0}^{64}\left(x^{1 / 3}-\left[x^{1 / 3}\right]\right) \mathrm{d} x$, then $\frac{1}{\pi} \int_{0}^{\alpha \pi}\left(\frac{\sin ^{2} \theta}{\sin ^{6} \theta+\cos ^{6} \theta}\right) \mathrm{d} \theta$ is equal to $\_\_\_\_$ .
Let $P_{1}: y=4 x^{2}$ and $P_{2}: y=x^{2}+27$ be two parabolas. If the area of the bounded region enclosed between $P_{1}$ and $P_{2}$ is six times the area of the bounded region enclosed between the line $y=\alpha x, \alpha>0$ and $P_{1}$, then $\alpha$ is equal to :
If $y=y(x)$ satisfies the differential equation $16(\sqrt{x+9 \sqrt{x}})(4+\sqrt{9+\sqrt{x}}) \cos y \mathrm{~d} y=(1+2 \sin y) \mathrm{d} x, x>0$ and $y(256)=\frac{\pi}{2}, y(49)=\alpha$, then $2 \sin \alpha$ is equal to :
Other Mathematics topics for JEE Main
Keep practicing Calculus
Work through every JEE Main Calculus PYQ, year by year.