What is the integral of x(x+3)^7 ?

The integral of x(x+3)^7 is 1/72 (x+3)^8(8x-3)+C

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integral of x(x+3)^7

\int x (x + 3)^{7} d x

\frac{1}{72} (x + 3)^{8} (8 x - 3) + C

Solution steps

\int x (x + 3)^{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}{8} u^{\frac{8}{7}})

Substitute back

u = (x + 3)^{7}

= \frac{1}{7} (\frac{7}{9} ((x + 3)^{7})^{\frac{9}{7}} - \frac{21}{8} ((x + 3)^{7})^{\frac{8}{7}})

Simplify

\frac{1}{7} (\frac{7}{9} ((x + 3)^{7})^{\frac{9}{7}} - \frac{21}{8} ((x + 3)^{7})^{\frac{8}{7}}) : \frac{1}{72} (x + 3)^{8} (8 x - 3)

= \frac{1}{72} (x + 3)^{8} (8 x - 3)

Add a constant to the solution

= \frac{1}{72} (x + 3)^{8} (8 x - 3) + C

x^{2} \frac{d y}{d x} = y (4 x + 3)

\int_{- 1}^{1} \frac{1}{x ^{2}} d x

\frac{\partial}{\partial y} (6 z e^{x yz})

\frac{1 + x}{1 - x}

\int_{- 1}^{2} \frac{4}{x ^{3}} d x

What is the integral of x(x+3)^7 ?
The integral of x(x+3)^7 is 1/72 (x+3)^8(8x-3)+C