The value of the integral ∫e^x2+1/x2x^2-1/x+1dx is equal to (where C is the constant of integration)

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

Q1 / 104 · NTA Abhyas (JEE Main) 2020 · Mathematics · Indefinite Integration

The value of the integral $\int e^{x^{2} + \frac{1}{x}} (2 x^{2} - \frac{1}{x} + 1) d x$ is equal to (where $C$ is the constant of integration)

A
$e^{x^{2} + \frac{1}{x}} + C$
B
$x^{2} e^{x^{2} + \frac{1}{x}} + C$
Correct answer
C
$x e^{x^{2} + \frac{1}{x}} + C$
D
$x \cdot e^{x^{2}} + C$

Solution

Ι = \int \underset{Ι}{\underset{⏟}{e^{x^{2} + \frac{1}{x}}}} \cdot \underset{Ι Ι}{\underset{⏟}{1}} d x + \int e^{(x^{2} + \frac{1}{x})} (2 x - \frac{1}{x^{2}}) \cdot x d x

Using integration by parts, we get,

Ι = e^{x^{2} + \frac{1}{x}} \cdot x - \int e^{(x^{2} + \frac{1}{x})} (2 x - \frac{1}{x^{2}}) \cdot x d x + \int e^{(x^{2} + \frac{1}{x})} (2 x - \frac{1}{x^{2}}) \cdot x d x

= x e^{x^{2} + \frac{1}{x}} + C

Hence, the correct answer is option: 'C'.

The value of the integral ∫ex2+1x2x2-1x+1dx is equal to (where C is the constant of integration)

Solution

More Indefinite Integration Questions

The value of the integral $\int e^{x^{2} + \frac{1}{x}} (2 x^{2} - \frac{1}{x} + 1) d x$ is equal to (where $C$ is the constant of integration)