What is the integral of+4x(x^2-3)^5 ?

The integral of+4x(x^2-3)^5 is 1/3 (x^2-3)^6+C

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integral of+4x(x^2-3)^5

\int + 4 x (x^{2} - 3)^{5} d x

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

Solution steps

\int 4 x (x^{2} - 3)^{5} d x

Take the constant out:

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

= 4 \cdot \int x (x^{2} - 3)^{5} d x

Apply u-substitution

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

= 4 \cdot \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

= 4 \cdot \frac{1}{2} \cdot \int u^{5} d u

Apply the Power Rule

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

= 4 \cdot \frac{1}{2} \cdot \frac{u ^{6}}{6}

Substitute back

u = x^{2} - 3

= 4 \cdot \frac{1}{2} \cdot \frac{( x ^{2} - 3 ) ^{6}}{6}

Simplify

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

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

Add a constant to the solution

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

\int (4 cos (x) + 5 sin (x)) d x

\frac{\partial}{\partial x} (x^{x^{2} - y^{2}})

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

\frac{d}{d x} (\frac{40}{3 x ^{\frac{1}{3}}})

x \to 0 lim (\frac{x + 4 - 2}{sin ( 5 x )})

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