What is the integral of (e^{2x}\sqrt[3]{e^x+2}) ?

The integral of (e^{2x}\sqrt[3]{e^x+2}) is 3/7 (e^x+2)^{7/3}-3/2 (e^x+2)^{4/3}+C

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integral of (e^{2x}\sqrt[3]{e^x+2})

\int (e^{2 x} 3 e^{x} + 2) d x

\frac{3}{7} (e^{x} + 2)^{\frac{7}{3}} - \frac{3}{2} (e^{x} + 2)^{\frac{4}{3}} + C

Solution steps

\int e^{2 x} 3 e^{x} + 2 d x

Apply u-substitution

: \int u 3 u + 2 d u

= \int u 3 u + 2 d u

Apply u-substitution

: \int (v - 2) 3 v d v

= \int (v - 2) 3 v d v

Expand

(v - 2) 3 v : v^{\frac{4}{3}} - 2 3 v

= \int v^{\frac{4}{3}} - 2 3 v d v

Apply the Sum Rule:

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

= \int v^{\frac{4}{3}} d v - \int 2 3 v d v

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

\int 2 3 v d v = \frac{3}{2} v^{\frac{4}{3}}

= \frac{3}{7} v^{\frac{7}{3}} - \frac{3}{2} v^{\frac{4}{3}}

Substitute back

= \frac{3}{7} (e^{x} + 2)^{\frac{7}{3}} - \frac{3}{2} (e^{x} + 2)^{\frac{4}{3}}

Add a constant to the solution

= \frac{3}{7} (e^{x} + 2)^{\frac{7}{3}} - \frac{3}{2} (e^{x} + 2)^{\frac{4}{3}} + C

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What is the integral of (e^{2x}\sqrt[3]{e^x+2}) ?
The integral of (e^{2x}\sqrt[3]{e^x+2}) is 3/7 (e^x+2)^{7/3}-3/2 (e^x+2)^{4/3}+C