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

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

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

\frac{1}{( 1 - t ^{2} ) ^{\frac{1}{2}}} + C

Solution steps

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

Apply u-substitution

: \int - \frac{1}{2 u ^{\frac{3}{2}}} d u

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

Apply the Power Rule

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

= - \frac{1}{2} (- \frac{2}{u ^{\frac{1}{2}}})

Substitute back

u = 1 - t^{2}

= - \frac{1}{2} (- \frac{2}{( 1 - t ^{2} ) ^{\frac{1}{2}}})

Simplify

- \frac{1}{2} (- \frac{2}{( 1 - t ^{2} ) ^{\frac{1}{2}}}) : \frac{1}{( 1 - t ^{2} ) ^{\frac{1}{2}}}

= \frac{1}{( 1 - t ^{2} ) ^{\frac{1}{2}}}

Add a constant to the solution

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

\frac{\partial}{\partial x} ((x - 1)^{2})

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

\int_{1}^{\infty} 10 e^{- 5 x} d x

\int_{0}^{2} 2 - x d x

10 e^{- 0.2 t}

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