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

The integral of (x^2-3)^5x is 1/12 (x^2-3)^6+C

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

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

\frac{1}{12} (x^{2} - 3)^{6} + C

Solution steps

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

Apply u-substitution

: \int \frac{u ^{5}}{2} d u

= \int \frac{u ^{5}}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int u^{5} d u

Apply the Power Rule

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

= \frac{1}{2} \cdot \frac{u ^{6}}{6}

Substitute back

u = x^{2} - 3

= \frac{1}{2} \cdot \frac{( x ^{2} - 3 ) ^{6}}{6}

Simplify

\frac{1}{2} \cdot \frac{( x ^{2} - 3 ) ^{6}}{6} : \frac{1}{12} (x^{2} - 3)^{6}

= \frac{1}{12} (x^{2} - 3)^{6}

Add a constant to the solution

= \frac{1}{12} (x^{2} - 3)^{6} + C

\int - 4 - x^{2} d x

(x^{2} + 2 y^{2}) = x y y^{^{'}}

y^{^{'}} = \frac{y ^{3} - 3}{( 3 x y ^{2} )}

f (x) = \frac{5 x}{e ^{x}}

\frac{1}{3 x ^{2} + 2 x}

What is the integral of (x^2-3)^5x ?
The integral of (x^2-3)^5x is 1/12 (x^2-3)^6+C