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

The integral of t/((2t^2-3)^{2/3)} is 3/4 (2t^2-3)^{1/3}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{3}{4} (2 t^{2} - 3)^{\frac{1}{3}} + C

Solution steps

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

Apply u-substitution

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

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

Apply u-substitution

: \int \frac{1}{2 v ^{\frac{2}{3}}} d v

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

Take the constant out:

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

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

Apply the Power Rule

: 3 v^{\frac{1}{3}}

= \frac{1}{2} \cdot \frac{1}{2} \cdot 3 v^{\frac{1}{3}}

Substitute back

= \frac{1}{2} \cdot \frac{1}{2} \cdot 3 (2 t^{2} - 3)^{\frac{1}{3}}

Simplify

\frac{1}{2} \cdot \frac{1}{2} \cdot 3 (2 t^{2} - 3)^{\frac{1}{3}} : \frac{3}{4} (2 t^{2} - 3)^{\frac{1}{3}}

= \frac{3}{4} (2 t^{2} - 3)^{\frac{1}{3}}

Add a constant to the solution

= \frac{3}{4} (2 t^{2} - 3)^{\frac{1}{3}} + C

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

\int z e^{x z} d x

\int x sin^{2} (ln (x)) d x

\int 3 x 2 x + 1 d x

\int \frac{8}{x} + 6 x d x

What is the integral of t/((2t^2-3)^{2/3)} ?
The integral of t/((2t^2-3)^{2/3)} is 3/4 (2t^2-3)^{1/3}+C