What is the integral of (5x^3+3x-8)^6(5x^2+1) ?

The integral of (5x^3+3x-8)^6(5x^2+1) is 1/21 (5x^3+3x-8)^7+C

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integral of (5x^3+3x-8)^6(5x^2+1)

\int (5 x^{3} + 3 x - 8)^{6} (5 x^{2} + 1) d x

\frac{1}{21} (5 x^{3} + 3 x - 8)^{7} + C

Solution steps

\int (5 x^{3} + 3 x - 8)^{6} (5 x^{2} + 1) d x

Apply u-substitution

= \int \frac{u ^{6}}{3} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} \cdot \int u^{6} d u

Apply the Power Rule

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

Substitute back

u = 5 x^{3} + 3 x - 8

= \frac{1}{3} \cdot \frac{( 5 x ^{3} + 3 x - 8 ) ^{7}}{7}

Simplify

\frac{1}{3} \cdot \frac{( 5 x ^{3} + 3 x - 8 ) ^{7}}{7} : \frac{1}{21} (5 x^{3} + 3 x - 8)^{7}

= \frac{1}{21} (5 x^{3} + 3 x - 8)^{7}

Add a constant to the solution

= \frac{1}{21} (5 x^{3} + 3 x - 8)^{7} + C

What is the integral of (5x^3+3x-8)^6(5x^2+1) ?
The integral of (5x^3+3x-8)^6(5x^2+1) is 1/21 (5x^3+3x-8)^7+C