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

The integral of (2t^2+1)^{1/3}t^3 is 1/8 (3/7 (2t^2+1)^{7/3}-3/4 (2t^2+1)^{4/3})+C

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

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

\frac{1}{8} (\frac{3}{7} (2 t^{2} + 1)^{\frac{7}{3}} - \frac{3}{4} (2 t^{2} + 1)^{\frac{4}{3}}) + C

Solution steps

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

Apply u-substitution

: \int \frac{u ^{\frac{1}{3}} ( u - 1 )}{8} d u

= \int \frac{u ^{\frac{1}{3}} ( u - 1 )}{8} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int u^{\frac{1}{3}} (u - 1) d u

Expand

u^{\frac{1}{3}} (u - 1) : u^{\frac{4}{3}} - u^{\frac{1}{3}}

= \frac{1}{8} \cdot \int u^{\frac{4}{3}} - u^{\frac{1}{3}} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{8} (\int u^{\frac{4}{3}} d u - \int u^{\frac{1}{3}} d u)

\int u^{\frac{4}{3}} d u = \frac{3}{7} u^{\frac{7}{3}}

\int u^{\frac{1}{3}} d u = \frac{3}{4} u^{\frac{4}{3}}

= \frac{1}{8} (\frac{3}{7} u^{\frac{7}{3}} - \frac{3}{4} u^{\frac{4}{3}})

Substitute back

u = 2 t^{2} + 1

= \frac{1}{8} (\frac{3}{7} (2 t^{2} + 1)^{\frac{7}{3}} - \frac{3}{4} (2 t^{2} + 1)^{\frac{4}{3}})

Add a constant to the solution

= \frac{1}{8} (\frac{3}{7} (2 t^{2} + 1)^{\frac{7}{3}} - \frac{3}{4} (2 t^{2} + 1)^{\frac{4}{3}}) + C

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What is the integral of (2t^2+1)^{1/3}t^3 ?
The integral of (2t^2+1)^{1/3}t^3 is 1/8 (3/7 (2t^2+1)^{7/3}-3/4 (2t^2+1)^{4/3})+C