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

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

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

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

Solution steps

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

Simplify

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

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

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

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

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

Apply u-substitution

: \int \frac{1}{2 u ^{\frac{1}{2}}} d u

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

Apply the Power Rule

: 2 u^{\frac{1}{2}}

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

Substitute back

u = 1 + x^{2}

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{x}{1 - a x ^{2}} d x

\int (3 x^{2} + 3) sin (3 x^{3} + 9 x) d x

\int - 8 e^{- 0.1 x} d x

\int \frac{1}{1 - \frac{x}{2}} d x

\int - 2 t e^{- t} d t

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