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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{u ^{\frac{3}{2}}}{2} d u

= \int \frac{u ^{\frac{3}{2}}}{2} d u

Take the constant out:

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

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

Apply the Power Rule

: \frac{2}{5} u^{\frac{5}{2}}

= \frac{1}{2} \cdot \frac{2}{5} u^{\frac{5}{2}}

Substitute back

u = 1 + x^{2}

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

Simplify

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

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

Add a constant to the solution

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

\int (\frac{1}{e ^{\frac{t}{4}}} - e^{\frac{t}{4}})^{2} d t

\int 10 - \frac{100}{x + 10} d x

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

\int cosh^{4} (x) sinh (x) d x

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

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