What is the integral of (2x^4+6)^{4/3}x^3 ?

The integral of (2x^4+6)^{4/3}x^3 is 3/56 (2x^4+6)^{7/3}+C

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integral of (2x^4+6)^{4/3}x^3

\int (2 x^{4} + 6)^{\frac{4}{3}} x^{3} d x

\frac{3}{56} (2 x^{4} + 6)^{\frac{7}{3}} + C

Solution steps

\int (2 x^{4} + 6)^{\frac{4}{3}} x^{3} d x

Apply u-substitution

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

Apply the Power Rule

= \frac{1}{8} \cdot \frac{3}{7} u^{\frac{7}{3}}

Substitute back

u = 2 x^{4} + 6

= \frac{1}{8} \cdot \frac{3}{7} (2 x^{4} + 6)^{\frac{7}{3}}

Simplify

\frac{1}{8} \cdot \frac{3}{7} (2 x^{4} + 6)^{\frac{7}{3}} : \frac{3}{56} (2 x^{4} + 6)^{\frac{7}{3}}

= \frac{3}{56} (2 x^{4} + 6)^{\frac{7}{3}}

Add a constant to the solution

= \frac{3}{56} (2 x^{4} + 6)^{\frac{7}{3}} + C

What is the integral of (2x^4+6)^{4/3}x^3 ?
The integral of (2x^4+6)^{4/3}x^3 is 3/56 (2x^4+6)^{7/3}+C