I=∫x4+x2x3−1dx
⇒I=∫x2+x12x−x21dx
Put, x2+x1=t⇒(2x−x21)dx=dt
∴I=∫tdt=loge(t)+C
=loge(x2+x1)+C
=loge∣x(x3+1)∣+C
JEE Main 2019 — Mathematics Calculus
The integral ∫x4+x2x3−1dx, is equal to
Held on 12 Apr 2019 · Verified 6 Jul 2026.
21loge∣x3∣(x3+1)2+C
logex2∣x3+1∣+C
loge∣x(x3+1)∣+C
21logex2∣x3+1∣+C
Sign in to track your attempts and accuracy.
Sign in to keep a private note on this question. Nothing you write is ever public.
Let $[\cdot]$ denote the greatest integer function. Then the value of $\displaystyle\int_0^3 \left(\dfrac{e^x + e^{-x}}{[x]!}\right) dx$ is :
The value of $\sum_{r=1}^{20}\left(\left|\sqrt{\pi\left(\int_{0}^{r} x|\sin \pi x| d x\right)}\right|\right)$ is $\_\_\_\_$
The value of ∫₀¹ x·eˣ dx is:
If the area of the region bounded by $16x^2 - 9y^2 = 144$ and $8x - 3y = 24$ is A, then $3(A + 6 \log_e(3))$ is equal to _______.
Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a differentiable function such that $f\left(\dfrac{x+y}{3}\right) = \dfrac{f(x) + f(y)}{3}$ for all $x, y \in \mathbb{R}$, and $f'(0) = 3$. Then the minimum value of the function $g(x) = 3 + e^x f(x)$, is:
Work through every JEE Main Calculus PYQ, year by year.