What is the integral of t(2t+7)^{1/3} ?

The integral of t(2t+7)^{1/3} is 1/4 (3/7 (2t+7)^{7/3}-21/4 (2t+7)^{4/3})+C

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integral of t(2t+7)^{1/3}

\int t (2 t + 7)^{\frac{1}{3}} d t

\frac{1}{4} (\frac{3}{7} (2 t + 7)^{\frac{7}{3}} - \frac{21}{4} (2 t + 7)^{\frac{4}{3}}) + C

Solution steps

\int t (2 t + 7)^{\frac{1}{3}} d t

Apply u-substitution

: \int \frac{u ^{\frac{1}{3}} ( u - 7 )}{4} d u

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

Take the constant out:

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

= \frac{1}{4} \cdot \int u^{\frac{1}{3}} (u - 7) d u

Expand

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

= \frac{1}{4} \cdot \int u^{\frac{4}{3}} - 7 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}{4} (\int u^{\frac{4}{3}} d u - \int 7 u^{\frac{1}{3}} d u)

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

\int 7 u^{\frac{1}{3}} d u = \frac{21}{4} u^{\frac{4}{3}}

= \frac{1}{4} (\frac{3}{7} u^{\frac{7}{3}} - \frac{21}{4} u^{\frac{4}{3}})

Substitute back

u = 2 t + 7

= \frac{1}{4} (\frac{3}{7} (2 t + 7)^{\frac{7}{3}} - \frac{21}{4} (2 t + 7)^{\frac{4}{3}})

Add a constant to the solution

= \frac{1}{4} (\frac{3}{7} (2 t + 7)^{\frac{7}{3}} - \frac{21}{4} (2 t + 7)^{\frac{4}{3}}) + C

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What is the integral of t(2t+7)^{1/3} ?
The integral of t(2t+7)^{1/3} is 1/4 (3/7 (2t+7)^{7/3}-21/4 (2t+7)^{4/3})+C