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

The integral of (2t^2+3)^{4/3}t^3 is 1/8 (3/10 (2t^2+3)^{10/3}-9/7 (2t^2+3)^{7/3})+C

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

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

\frac{1}{8} (\frac{3}{10} (2 t^{2} + 3)^{\frac{10}{3}} - \frac{9}{7} (2 t^{2} + 3)^{\frac{7}{3}}) + C

Solution steps

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

Apply u-substitution

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

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

Expand

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

= \frac{1}{8} \cdot \int u^{\frac{7}{3}} - 3 u^{\frac{4}{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{7}{3}} d u - \int 3 u^{\frac{4}{3}} d u)

\int u^{\frac{7}{3}} d u = \frac{3}{10} u^{\frac{10}{3}}

\int 3 u^{\frac{4}{3}} d u = \frac{9}{7} u^{\frac{7}{3}}

= \frac{1}{8} (\frac{3}{10} u^{\frac{10}{3}} - \frac{9}{7} u^{\frac{7}{3}})

Substitute back

u = 2 t^{2} + 3

= \frac{1}{8} (\frac{3}{10} (2 t^{2} + 3)^{\frac{10}{3}} - \frac{9}{7} (2 t^{2} + 3)^{\frac{7}{3}})

Add a constant to the solution

= \frac{1}{8} (\frac{3}{10} (2 t^{2} + 3)^{\frac{10}{3}} - \frac{9}{7} (2 t^{2} + 3)^{\frac{7}{3}}) + C

f (x) = (x^{2} + 3 x - 1) e^{x}

f (x) = - x^{2} + 5 x

- 3 lo g_{4} (x)

n \to \infty lim (\frac{( 2 n ) ^{2 n}}{( 2 n + 2 ) ^{2 n}})

x \to 2 lim (3 x - 1)

What is the integral of (2t^2+3)^{4/3}t^3 ?
The integral of (2t^2+3)^{4/3}t^3 is 1/8 (3/10 (2t^2+3)^{10/3}-9/7 (2t^2+3)^{7/3})+C