What is the integral of p(p+8)^4 ?

The integral of p(p+8)^4 is (p^6)/6+(32p^5)/5+96p^4+(2048p^3)/3+2048p^2+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of p(p+8)^4

\int p (p + 8)^{4} d p

\frac{p ^{6}}{6} + \frac{32 p ^{5}}{5} + 96 p^{4} + \frac{2048 p ^{3}}{3} + 2048 p^{2} + C

Solution steps

\int p (p + 8)^{4} d p

Expand

p (p + 8)^{4} : p^{5} + 32 p^{4} + 384 p^{3} + 2048 p^{2} + 4096 p

= \int p^{5} + 32 p^{4} + 384 p^{3} + 2048 p^{2} + 4096 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^{5} d p + \int 32 p^{4} d p + \int 384 p^{3} d p + \int 2048 p^{2} d p + \int 4096 p d p

\int p^{5} d p = \frac{p ^{6}}{6}

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

\int 384 p^{3} d p = 96 p^{4}

\int 2048 p^{2} d p = \frac{2048 p ^{3}}{3}

\int 4096 p d p = 2048 p^{2}

= \frac{p ^{6}}{6} + \frac{32 p ^{5}}{5} + 96 p^{4} + \frac{2048 p ^{3}}{3} + 2048 p^{2}

Add a constant to the solution

= \frac{p ^{6}}{6} + \frac{32 p ^{5}}{5} + 96 p^{4} + \frac{2048 p ^{3}}{3} + 2048 p^{2} + C

What is the integral of p(p+8)^4 ?
The integral of p(p+8)^4 is (p^6)/6+(32p^5)/5+96p^4+(2048p^3)/3+2048p^2+C