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

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

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

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

Solution steps

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

Apply Trigonometric Substitution

: \int cos (u) d u

= \int cos (u) d u

Use the common integral:

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

= sin (u)

Substitute back

u = arctan (x)

= sin (arctan (x))

Use the following identity:

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

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

Add a constant to the solution

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

\int \frac{1}{x + 1 - x - 1} d x

\int (x^{4} - 6 x^{2} + 3) d x

\int \frac{3 x ^{4} - 12}{x ^{2} + 2} d x

\int x^{2} - 2 x - 3 d x

\int arcsin (\frac{x}{1 + x}) d x

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