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

The integral of (x^2+1)(x^3+3x-1)^5 is 1/18 (x^3+3x-1)^6+C

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

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

\frac{1}{18} (x^{3} + 3 x - 1)^{6} + C

Solution steps

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

Apply u-substitution

: \int \frac{u ^{5}}{3} d u

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

Apply the Power Rule

: \frac{u ^{6}}{6}

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

Substitute back

u = x^{3} + 3 x - 1

= \frac{1}{3} \cdot \frac{( x ^{3} + 3 x - 1 ) ^{6}}{6}

Simplify

\frac{1}{3} \cdot \frac{( x ^{3} + 3 x - 1 ) ^{6}}{6} : \frac{1}{18} (x^{3} + 3 x - 1)^{6}

= \frac{1}{18} (x^{3} + 3 x - 1)^{6}

Add a constant to the solution

= \frac{1}{18} (x^{3} + 3 x - 1)^{6} + C

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What is the integral of (x^2+1)(x^3+3x-1)^5 ?
The integral of (x^2+1)(x^3+3x-1)^5 is 1/18 (x^3+3x-1)^6+C