What is the integral of 3/(x^2sqrt(1+x^2)) ?

The integral of 3/(x^2sqrt(1+x^2)) is -(3sqrt(1+x^2))/x+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 3/(x^2sqrt(1+x^2))

\int \frac{3}{x ^{2} 1 + x ^{2}} d x

- \frac{3 1 + x ^{2}}{x} + C

Solution steps

\int \frac{3}{x ^{2} 1 + x ^{2}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 \cdot \int \frac{1}{x ^{2} 1 + x ^{2}} d x

Apply Trigonometric Substitution

: \int cot^{2} (u) sec (u) d u

= 3 \cdot \int cot^{2} (u) sec (u) d u

Simplify

cot^{2} (u) sec (u) : cot (u) csc (u)

= 3 \cdot \int cot (u) csc (u) d u

Apply u-substitution

: \int - 1 d v

= 3 \cdot \int - 1 d v

Integral of a constant:

\int a d x = a x

= 3 (- 1) v

Substitute back

= 3 (- 1) csc (arctan (x))

Simplify

3 (- 1) csc (arctan (x)) : - \frac{3 1 + x ^{2}}{x}

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

Add a constant to the solution

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

\frac{\partial}{\partial y} (x^{2} + y^{2})

\int_{0}^{8} \frac{5}{x ^{\frac{2}{3}}} d x

y = 2 x^{2} + 1

\frac{d}{d x} ((cos (3 x))^{x})

What is the integral of 3/(x^2sqrt(1+x^2)) ?
The integral of 3/(x^2sqrt(1+x^2)) is -(3sqrt(1+x^2))/x+C