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

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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

= \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{u ^{10}}{10}

= \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

\frac{d y}{d x} = \frac{y}{1 + x ^{2}}

f^{^{''}} (x) = \frac{3}{4} x^{4} + 2 x^{3} - 12 x^{2} + 29

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

\frac{( 2 x ^{2} tan ( x ) )}{sec ( 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