What is the integral of (2x+7)(x^2+7x+3)^{4/5} ?

The integral of (2x+7)(x^2+7x+3)^{4/5} is 5/9 (x^2+7x+3)^{9/5}+C

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integral of (2x+7)(x^2+7x+3)^{4/5}

\int (2 x + 7) (x^{2} + 7 x + 3)^{\frac{4}{5}} d x

\frac{5}{9} (x^{2} + 7 x + 3)^{\frac{9}{5}} + C

Solution steps

\int (2 x + 7) (x^{2} + 7 x + 3)^{\frac{4}{5}} d x

Apply u-substitution

= \int u^{\frac{4}{5}} d u

Apply the Power Rule

= \frac{5}{9} u^{\frac{9}{5}}

Substitute back

u = x^{2} + 7 x + 3

= \frac{5}{9} (x^{2} + 7 x + 3)^{\frac{9}{5}}

Add a constant to the solution

= \frac{5}{9} (x^{2} + 7 x + 3)^{\frac{9}{5}} + C

What is the integral of (2x+7)(x^2+7x+3)^{4/5} ?
The integral of (2x+7)(x^2+7x+3)^{4/5} is 5/9 (x^2+7x+3)^{9/5}+C