The value of the integral ∫ e^ x^ 2 + (1)/(x) (2x^ 2 - (1)/(x) + 1)dx is equal to (where C is the constant of integration)

Correct answer: C. xe^ x^ 2 + (1)/(x) + C Ι = ∫ e^ x^ 2 + (1)/(x) ⏟ _ Ι ⋅ 1 ⏟ _ ΙΙ dx + ∫ e(x^ 2 + (1)/(x))(2x - (1)/(x2)) ⋅ xdx Using integration by parts, we get, Ι = e^ x^ 2 + (1)/(x) · x - ∫ e(x^ 2 + (1)/(x))(2x - (1)/(x2)) ⋅ xdx + ∫ e(x^ 2 + (1)/(x))(2x - (1)/(x2)) ⋅ xdx = xe^ x^ 2 + (1)/(x) + C

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Indefinite Integration

NTA Abhyas 2020 · Mathematics · Indefinite Integration

The value of the integral ∫ e^ x^ 2 + (1)/(x) (2x^ 2 - (1)/(x) +...

The value of the integral

∫ e^{x^{2} + \dfrac{1}{x}}

\left(2x^{2} - \dfrac{1}{x} + 1\right)dx

is equal to (where

C is the constant of integration)

Options

A $e^{x^{2} + \dfrac{1}{x}} + C$
B $x^{2}e^{x^{2} + \dfrac{1}{x}} + C$
C $xe^{x^{2} + \dfrac{1}{x}} + C$
D $x ⋅ e^{x^{2}} + C$

Correct answer

C. $xe^{x^{2} + \dfrac{1}{x}} + C$

Step-by-step solution

Ι = ∫ \mathop{e^{x^{2} + \dfrac{1}{x}} ⏟}_{Ι} ⋅ \mathop{1 ⏟}_{ΙΙ}dx + ∫ e\left(x^{2} + \dfrac{1}{x}\right)\left(2x - \dfrac{1}{x2}\right) ⋅ xdx

Using integration by parts, we get,

Ι = e^{x^{2} + \dfrac{1}{x}} \cdot x - ∫ e\left(x^{2} + \dfrac{1}{x}\right)\left(2x - \dfrac{1}{x2}\right) ⋅ xdx + ∫ e\left(x^{2} + \dfrac{1}{x}\right)\left(2x - \dfrac{1}{x2}\right) ⋅ xdx

= xe^{x^{2} + \dfrac{1}{x}} + C

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