What is the integral of (2x^7+3x-9)^{1/4}(14x^6+3) ?

The integral of (2x^7+3x-9)^{1/4}(14x^6+3) is 4/5 (2x^7+3x-9)^{5/4}+C

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integral of (2x^7+3x-9)^{1/4}(14x^6+3)

\int (2 x^{7} + 3 x - 9)^{\frac{1}{4}} (14 x^{6} + 3) d x

\frac{4}{5} (2 x^{7} + 3 x - 9)^{\frac{5}{4}} + C

Solution steps

\int (2 x^{7} + 3 x - 9)^{\frac{1}{4}} (14 x^{6} + 3) d x

Apply u-substitution

: \int u^{\frac{1}{4}} d u

= \int u^{\frac{1}{4}} d u

Apply the Power Rule

: \frac{4}{5} u^{\frac{5}{4}}

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

Substitute back

u = 2 x^{7} + 3 x - 9

= \frac{4}{5} (2 x^{7} + 3 x - 9)^{\frac{5}{4}}

Add a constant to the solution

= \frac{4}{5} (2 x^{7} + 3 x - 9)^{\frac{5}{4}} + C

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What is the integral of (2x^7+3x-9)^{1/4}(14x^6+3) ?
The integral of (2x^7+3x-9)^{1/4}(14x^6+3) is 4/5 (2x^7+3x-9)^{5/4}+C