What is the integral of 8/((x^2+1)^{3/2)} ?

The integral of 8/((x^2+1)^{3/2)} is (8x)/((1+x^2)^{1/2)}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 8/((x^2+1)^{3/2)}

\int \frac{8}{( x ^{2} + 1 ) ^{\frac{3}{2}}} d x

\frac{8 x}{( 1 + x ^{2} ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{8}{( x ^{2} + 1 ) ^{\frac{3}{2}}} d x

Take the constant out:

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

= 8 \cdot \int \frac{1}{( x ^{2} + 1 ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

: \int cos (u) d u

= 8 \cdot \int cos (u) d u

Use the common integral:

\int cos (u) d u = sin (u)

= 8 sin (u)

Substitute back

u = arctan (x)

= 8 sin (arctan (x))

Simplify

8 sin (arctan (x)) : \frac{8 x}{( 1 + x ^{2} ) ^{\frac{1}{2}}}

= \frac{8 x}{( 1 + x ^{2} ) ^{\frac{1}{2}}}

Add a constant to the solution

= \frac{8 x}{( 1 + x ^{2} ) ^{\frac{1}{2}}} + C

x^{2} \frac{d y}{d x} + 2 x y = 5 y^{4}

\int \frac{x}{4} d x

\frac{\partial}{\partial y} (\frac{x}{y})

x \to 2 lim (\frac{( x - 2 )}{x})

(((- 3)^{^{'}})^{4})^{^{'}}

What is the integral of 8/((x^2+1)^{3/2)} ?
The integral of 8/((x^2+1)^{3/2)} is (8x)/((1+x^2)^{1/2)}+C