What is the integral of (3x)/((x^2+1)^{3/5)} ?

The integral of (3x)/((x^2+1)^{3/5)} is 15/4 (x^2+1)^{2/5}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (3x)/((x^2+1)^{3/5)}

\int \frac{3 x}{( x ^{2} + 1 ) ^{\frac{3}{5}}} d x

\frac{15}{4} (x^{2} + 1)^{\frac{2}{5}} + C

Solution steps

\int \frac{3 x}{( x ^{2} + 1 ) ^{\frac{3}{5}}} d x

Take the constant out:

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

= 3 \cdot \int \frac{x}{( x ^{2} + 1 ) ^{\frac{3}{5}}} d x

Apply u-substitution

= 3 \cdot \int \frac{1}{2 ( u + 1 ) ^{\frac{3}{5}}} d u

Take the constant out:

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

= 3 \cdot \frac{1}{2} \cdot \int \frac{1}{( u + 1 ) ^{\frac{3}{5}}} d u

Apply u-substitution

= 3 \cdot \frac{1}{2} \cdot \int \frac{1}{v ^{\frac{3}{5}}} d v

Apply the Power Rule

= 3 \cdot \frac{1}{2} \cdot \frac{5}{2} v^{\frac{2}{5}}

Substitute back

= 3 \cdot \frac{1}{2} \cdot \frac{5}{2} (x^{2} + 1)^{\frac{2}{5}}

Simplify

3 \cdot \frac{1}{2} \cdot \frac{5}{2} (x^{2} + 1)^{\frac{2}{5}} : \frac{15}{4} (x^{2} + 1)^{\frac{2}{5}}

= \frac{15}{4} (x^{2} + 1)^{\frac{2}{5}}

Add a constant to the solution

= \frac{15}{4} (x^{2} + 1)^{\frac{2}{5}} + C

What is the integral of (3x)/((x^2+1)^{3/5)} ?
The integral of (3x)/((x^2+1)^{3/5)} is 15/4 (x^2+1)^{2/5}+C