What is the integral of p(p+4)^6 ?

The integral of p(p+4)^6 is (p^8)/8+(24p^7)/7+40p^6+256p^5+960p^4+2048p^3+2048p^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)^6

\int p (p + 4)^{6} d p

\frac{p ^{8}}{8} + \frac{24 p ^{7}}{7} + 40 p^{6} + 256 p^{5} + 960 p^{4} + 2048 p^{3} + 2048 p^{2} + C

Solution steps

\int p (p + 4)^{6} d p

Expand

p (p + 4)^{6} : p^{7} + 24 p^{6} + 240 p^{5} + 1280 p^{4} + 3840 p^{3} + 6144 p^{2} + 4096 p

= \int p^{7} + 24 p^{6} + 240 p^{5} + 1280 p^{4} + 3840 p^{3} + 6144 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^{7} d p + \int 24 p^{6} d p + \int 240 p^{5} d p + \int 1280 p^{4} d p + \int 3840 p^{3} d p + \int 6144 p^{2} d p + \int 4096 p d p

\int p^{7} d p = \frac{p ^{8}}{8}

\int 24 p^{6} d p = \frac{24 p ^{7}}{7}

\int 240 p^{5} d p = 40 p^{6}

\int 1280 p^{4} d p = 256 p^{5}

\int 3840 p^{3} d p = 960 p^{4}

\int 6144 p^{2} d p = 2048 p^{3}

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

= \frac{p ^{8}}{8} + \frac{24 p ^{7}}{7} + 40 p^{6} + 256 p^{5} + 960 p^{4} + 2048 p^{3} + 2048 p^{2}

Add a constant to the solution

= \frac{p ^{8}}{8} + \frac{24 p ^{7}}{7} + 40 p^{6} + 256 p^{5} + 960 p^{4} + 2048 p^{3} + 2048 p^{2} + C

\int lo g_{3} (2 x + 3) d x

\int - sin (x) sec^{2} (x) d x

\int (tan (3 x)) d x

\int \frac{sec ( θ )}{sec ( θ ) - cos ( θ )} d θ

\int \frac{1}{x ^{6}} d x

What is the integral of p(p+4)^6 ?
The integral of p(p+4)^6 is (p^8)/8+(24p^7)/7+40p^6+256p^5+960p^4+2048p^3+2048p^2+C