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

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

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

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{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

\frac{d}{d x} (3 (x^{2} + y^{2})^{2})

d er i v a t i v e y = 4 cot (3 - 2 x)

\int_{0}^{2} (4 x - 2 x^{2}) d x

\frac{d y}{d x} y = (x^{2} + 3 x)^{2}

x \to - \infty lim (\frac{4}{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