What is the integral of x(1+x)^6 ?

The integral of x(1+x)^6 is (x^2)/2+2x^3+(15x^4)/4+4x^5+(5x^6)/2+(6x^7)/7+(x^8)/8+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of x(1+x)^6

\int x (1 + x)^{6} d x

\frac{x ^{2}}{2} + 2 x^{3} + \frac{15 x ^{4}}{4} + 4 x^{5} + \frac{5 x ^{6}}{2} + \frac{6 x ^{7}}{7} + \frac{x ^{8}}{8} + C

Solution steps

\int x (1 + x)^{6} d x

Expand

x (1 + x)^{6} : x + 6 x^{2} + 15 x^{3} + 20 x^{4} + 15 x^{5} + 6 x^{6} + x^{7}

= \int x + 6 x^{2} + 15 x^{3} + 20 x^{4} + 15 x^{5} + 6 x^{6} + x^{7} d x

Apply the Sum Rule:

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

= \int x d x + \int 6 x^{2} d x + \int 15 x^{3} d x + \int 20 x^{4} d x + \int 15 x^{5} d x + \int 6 x^{6} d x + \int x^{7} d x

\int x d x = \frac{x ^{2}}{2}

\int 6 x^{2} d x = 2 x^{3}

\int 15 x^{3} d x = \frac{15 x ^{4}}{4}

\int 20 x^{4} d x = 4 x^{5}

\int 15 x^{5} d x = \frac{5 x ^{6}}{2}

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

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

= \frac{x ^{2}}{2} + 2 x^{3} + \frac{15 x ^{4}}{4} + 4 x^{5} + \frac{5 x ^{6}}{2} + \frac{6 x ^{7}}{7} + \frac{x ^{8}}{8}

Add a constant to the solution

= \frac{x ^{2}}{2} + 2 x^{3} + \frac{15 x ^{4}}{4} + 4 x^{5} + \frac{5 x ^{6}}{2} + \frac{6 x ^{7}}{7} + \frac{x ^{8}}{8} + C

\int \frac{cos ( ln ( t ) )}{2 t} d t

\int \frac{5 x - 4}{x ( x ^{2} - 2 x + 2 )} d x

\int \frac{x ^{3}}{1 - x} d x

\int 5 e^{5} x d x

\int x e^{x^{2}} sin (x^{2} + 4) d x

What is the integral of x(1+x)^6 ?
The integral of x(1+x)^6 is (x^2)/2+2x^3+(15x^4)/4+4x^5+(5x^6)/2+(6x^7)/7+(x^8)/8+C