((1+x+x^2)/(1+x^2)) e^(^(-1) x) dx = Option 1: x + e^(^(-1) x) + c Option 2: e^(^(-1) x) - x + c Option 3: e^(^(-1) x) + c Option 4: x e^(^(-1) x) + c

Answer: x e^(^(-1) x) + c. Solution: Note that d/dx[x * e^(^(-1)x)] = e^(^(-1)x) + (x * e^(^(-1)x))/(1+x^2) = e^(^(-1)x) * (1+x^2+x)/(1+x^2). So the integral equals x * e^(^(-1)x) + c.

Calculus

Integrals

Hard

core

$\int \left(\frac{1+x+x^2}{1+x^2}\right) e^{\tan^{-1} x} dx =$

$x + e^{\tan^{-1} x} + c$

$e^{\tan^{-1} x} - x + c$

$e^{\tan^{-1} x} + c$

$x e^{\tan^{-1} x} + c$

✅ Correct Option: 4

Note that $\frac{d}{dx}[x \cdot e^{\tan^{-1}x}] = e^{\tan^{-1}x} + \frac{x \cdot e^{\tan^{-1}x}}{1+x^2} = e^{\tan^{-1}x} \cdot \frac{1+x^2+x}{1+x^2}$ . So the integral equals $x \cdot e^{\tan^{-1}x} + c$ .

2023: 23 May Shift 3

The value of integral $\int \sqrt{4x^2 + 9}\, dx$ is

2025: 21 May Shift 1

Match List-I with List-II

List-I	List-II
Definite integral	Value
---	---
(A) $\int_{1}^{e} \frac{\log x}{x} dx$	(I) 4
(B) $\int_{-2}^{2} x^3(1 - x^2) dx$	(II) $\frac{1}{2}$
(C) $\int_{1}^{2} x \, dx$	(III) 0
(D) $\int_{-2}^{2} \vert x\vert dx$	(IV) $\frac{3}{2}$

Choose the correct answer from the options given below:

2022: 23 Aug Shift 1

$\int \sqrt{x^2 - 4x + 5} \, dx =$

2023: 23 May Shift 3

$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{dx}{1 + \sqrt{\tan x}}$ is equal to :

2026: 19 May Shift 2

$\int \frac{\log x - 3}{(\log x)^4} dx$ is equal to

(where C is an arbitrary constant and $\log_e x = \log x$ )

2026: 14 May Shift 2

If $\int_0^1 \frac{1-x}{1+x} dx = A\log 2 + B$ , then the value of A and B are

∫(1+x+x21+x2)etan⁡−1xdx=\int \left(\frac{1+x+x^2}{1+x^2}\right) e^{\tan^{-1} x} dx =∫(1+x21+x+x2​)etan−1xdx=

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

$\int \left(\frac{1+x+x^2}{1+x^2}\right) e^{\tan^{-1} x} dx =$