What is the integral of (4t^3+1)^{5/3}t^2 ?

The integral of (4t^3+1)^{5/3}t^2 is 1/32 (4t^3+1)^{8/3}+C

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integral of (4t^3+1)^{5/3}t^2

\int (4 t^{3} + 1)^{\frac{5}{3}} t^{2} d t

\frac{1}{32} (4 t^{3} + 1)^{\frac{8}{3}} + C

Solution steps

\int (4 t^{3} + 1)^{\frac{5}{3}} t^{2} d t

Apply u-substitution

: \int \frac{( 4 u + 1 ) ^{\frac{5}{3}}}{3} d u

= \int \frac{( 4 u + 1 ) ^{\frac{5}{3}}}{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 (4 u + 1)^{\frac{5}{3}} d u

Apply u-substitution

: \int \frac{v ^{\frac{5}{3}}}{4} d v

= \frac{1}{3} \cdot \int \frac{v ^{\frac{5}{3}}}{4} 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}{4} \cdot \int v^{\frac{5}{3}} d v

Apply the Power Rule

: \frac{3}{8} v^{\frac{8}{3}}

= \frac{1}{3} \cdot \frac{1}{4} \cdot \frac{3}{8} v^{\frac{8}{3}}

Substitute back

= \frac{1}{3} \cdot \frac{1}{4} \cdot \frac{3}{8} (4 t^{3} + 1)^{\frac{8}{3}}

Simplify

\frac{1}{3} \cdot \frac{1}{4} \cdot \frac{3}{8} (4 t^{3} + 1)^{\frac{8}{3}} : \frac{1}{32} (4 t^{3} + 1)^{\frac{8}{3}}

= \frac{1}{32} (4 t^{3} + 1)^{\frac{8}{3}}

Add a constant to the solution

= \frac{1}{32} (4 t^{3} + 1)^{\frac{8}{3}} + C

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What is the integral of (4t^3+1)^{5/3}t^2 ?
The integral of (4t^3+1)^{5/3}t^2 is 1/32 (4t^3+1)^{8/3}+C