What is the integral of p(p+5)^7 ?

The integral of p(p+5)^7 is 1/72 (p+5)^8(8p-5)+C

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integral of p(p+5)^7

\int p (p + 5)^{7} d p

\frac{1}{72} (p + 5)^{8} (8 p - 5) + C

Solution steps

\int p (p + 5)^{7} d p

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{35}{8} u^{\frac{8}{7}})

Substitute back

u = (p + 5)^{7}

= \frac{1}{7} (\frac{7}{9} ((p + 5)^{7})^{\frac{9}{7}} - \frac{35}{8} ((p + 5)^{7})^{\frac{8}{7}})

Simplify

\frac{1}{7} (\frac{7}{9} ((p + 5)^{7})^{\frac{9}{7}} - \frac{35}{8} ((p + 5)^{7})^{\frac{8}{7}}) : \frac{1}{72} (p + 5)^{8} (8 p - 5)

= \frac{1}{72} (p + 5)^{8} (8 p - 5)

Add a constant to the solution

= \frac{1}{72} (p + 5)^{8} (8 p - 5) + C

\int \frac{x}{x ^{2} + 6 x + 10} d x

\frac{d}{d x} (\frac{6 x ^{2} + 6 x + 8}{x})

y^{^{'}} + 6 (tan (6 x)) y = - cos (6 x)

\int 2^{x} (1 + 5^{x}) d x

\frac{d y}{d t} = y^{\frac{1}{3}}, y (0) = 0

What is the integral of p(p+5)^7 ?
The integral of p(p+5)^7 is 1/72 (p+5)^8(8p-5)+C