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

The integral of x(x^2+4)^5 is 1/12 (x^2+4)^6+C

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

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

\frac{1}{12} (x^{2} + 4)^{6} + C

Solution steps

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

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

Simplify

\frac{1}{2} \cdot \frac{( x ^{2} + 4 ) ^{6}}{6} : \frac{1}{12} (x^{2} + 4)^{6}

= \frac{1}{12} (x^{2} + 4)^{6}

Add a constant to the solution

= \frac{1}{12} (x^{2} + 4)^{6} + C

\frac{\partial}{\partial x} (e^{y x})

\int \frac{e ^{2 y}}{e ^{y} - 1} d y

\int sec^{3} (5 x) d x

y = arcsin (x)

\int \frac{x}{( x ^{2} - 1 ) ^{\frac{1}{2}}} d x

What is the integral of x(x^2+4)^5 ?
The integral of x(x^2+4)^5 is 1/12 (x^2+4)^6+C