What is the integral of x(4x^2+7)^9 ?

The integral of x(4x^2+7)^9 is 1/80 (4x^2+7)^{10}+C

integral of x(4x^2+7)^9

\int x (4 x^{2} + 7)^{9} d x

\frac{1}{80} (4 x^{2} + 7)^{10} + C

Solution steps

\int x (4 x^{2} + 7)^{9} d x

Apply u-substitution

= \int \frac{u ^{9}}{8} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int u^{9} d u

Apply the Power Rule

= \frac{1}{8} \cdot \frac{u ^{10}}{10}

Substitute back

u = 4 x^{2} + 7

= \frac{1}{8} \cdot \frac{( 4 x ^{2} + 7 ) ^{10}}{10}

Simplify

\frac{1}{8} \cdot \frac{( 4 x ^{2} + 7 ) ^{10}}{10} : \frac{1}{80} (4 x^{2} + 7)^{10}

= \frac{1}{80} (4 x^{2} + 7)^{10}

Add a constant to the solution

= \frac{1}{80} (4 x^{2} + 7)^{10} + C

\int \frac{1}{x \cdot ( 2 + 4 x )} d x

\frac{\partial}{\partial y} (x + 4 y)

d er i v a t i v e \frac{( 2 x ^{2} tan ( x ) )}{sec ( x )}

x \to 0 lim (e^{x} cos (2 x))

\int_{2}^{e^{3}} \frac{1}{x} d x

What is the integral of x(4x^2+7)^9 ?
The integral of x(4x^2+7)^9 is 1/80 (4x^2+7)^{10}+C