What is the integral of x(x+6)^7 ?

The integral of x(x+6)^7 is 1/36 (x+6)^8(4x-3)+C

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integral of x(x+6)^7

\int x (x + 6)^{7} d x

\frac{1}{36} (x + 6)^{8} (4 x - 3) + C

Solution steps

\int x (x + 6)^{7} d x

Apply u-substitution

Take the constant out:

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

Expand

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

\int u^{\frac{2}{7}} d u = \frac{7}{9} u^{\frac{9}{7}}

= \frac{1}{7} (\frac{7}{9} u^{\frac{9}{7}} - \frac{21}{4} u^{\frac{8}{7}})

Substitute back

u = (x + 6)^{7}

= \frac{1}{7} (\frac{7}{9} ((x + 6)^{7})^{\frac{9}{7}} - \frac{21}{4} ((x + 6)^{7})^{\frac{8}{7}})

Simplify

\frac{1}{7} (\frac{7}{9} ((x + 6)^{7})^{\frac{9}{7}} - \frac{21}{4} ((x + 6)^{7})^{\frac{8}{7}}) : \frac{1}{36} (x + 6)^{8} (4 x - 3)

= \frac{1}{36} (x + 6)^{8} (4 x - 3)

Add a constant to the solution

= \frac{1}{36} (x + 6)^{8} (4 x - 3) + C

\frac{\partial}{\partial y} (cos (x) + x cos (y) - y)

n = 0 \sum \infty (\frac{2}{4})^{n}

\int 5 x + 12 d x

n = 0 \sum \infty n \cdot (\frac{1}{6})^{n}

x y^{^{'}} + 2 y - 2 x^{2} y^{2} = 0

What is the integral of x(x+6)^7 ?
The integral of x(x+6)^7 is 1/36 (x+6)^8(4x-3)+C