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

The integral of p(p+4)^5 is (p^7)/7+(10p^6)/3+32p^5+160p^4+(1280p^3)/3+512p^2+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of p(p+4)^5

\int p (p + 4)^{5} d p

\frac{p ^{7}}{7} + \frac{10 p ^{6}}{3} + 32 p^{5} + 160 p^{4} + \frac{1280 p ^{3}}{3} + 512 p^{2} + C

Solution steps

\int p (p + 4)^{5} d p

Expand

p (p + 4)^{5} : p^{6} + 20 p^{5} + 160 p^{4} + 640 p^{3} + 1280 p^{2} + 1024 p

= \int p^{6} + 20 p^{5} + 160 p^{4} + 640 p^{3} + 1280 p^{2} + 1024 p d p

Apply the Sum Rule:

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

= \int p^{6} d p + \int 20 p^{5} d p + \int 160 p^{4} d p + \int 640 p^{3} d p + \int 1280 p^{2} d p + \int 1024 p d p

\int p^{6} d p = \frac{p ^{7}}{7}

\int 20 p^{5} d p = \frac{10 p ^{6}}{3}

\int 160 p^{4} d p = 32 p^{5}

\int 640 p^{3} d p = 160 p^{4}

\int 1280 p^{2} d p = \frac{1280 p ^{3}}{3}

\int 1024 p d p = 512 p^{2}

= \frac{p ^{7}}{7} + \frac{10 p ^{6}}{3} + 32 p^{5} + 160 p^{4} + \frac{1280 p ^{3}}{3} + 512 p^{2}

Add a constant to the solution

= \frac{p ^{7}}{7} + \frac{10 p ^{6}}{3} + 32 p^{5} + 160 p^{4} + \frac{1280 p ^{3}}{3} + 512 p^{2} + C

\frac{d}{d x} (\frac{e ^{x}}{5 + e ^{x}})

\frac{\partial}{\partial y} (2 x - 5 y + 3)

\int (1 + e^{x} y + x e^{x} y) d x

\int_{1}^{\infty} x e^{- 2 x} d x

\frac{d}{d x} (a x^{2} e^{- 4 x})

What is the integral of p(p+4)^5 ?
The integral of p(p+4)^5 is (p^7)/7+(10p^6)/3+32p^5+160p^4+(1280p^3)/3+512p^2+C