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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Apply u-substitution

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

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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