$\int\left(x+\frac{1}{x}\right)^2 dx$ eq

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Target Exam

CUET

Subject

-- Mathematics - Section A

Chapter

Determinants

Question:

$\int\left(x+\frac{1}{x}\right)^2 dx$ equals :

Options:

$\frac{x^3}{3}+\frac{1}{x}-2 x+c$

$\frac{x^3}{3}-\frac{1}{x}+2 x+c$

$\frac{x^3}{3}-\frac{1}{x}-2 x+c$

$\frac{x^3}{3}+\frac{1}{x}+2 x+c$

Correct Answer:

$\frac{x^3}{3}-\frac{1}{x}+2 x+c$

Explanation:

$I=\int\left(x+\frac{1}{x}\right)^2 d x$

so $I=\int x^2+\frac{1}{x^2}+2 d x$

$=\frac{x^3}{3}-\frac{1}{x}+2 x+c$