What is the integral of x^3(1+x^4)^3 ?

The integral of x^3(1+x^4)^3 is (x^4)/4+(3x^8)/8+(x^{12})/4+(x^{16})/(16)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of x^3(1+x^4)^3

\int x^{3} (1 + x^{4})^{3} d x

\frac{x ^{4}}{4} + \frac{3 x ^{8}}{8} + \frac{x ^{12}}{4} + \frac{x ^{16}}{16} + C

Solution steps

\int x^{3} (1 + x^{4})^{3} d x

Expand

x^{3} (1 + x^{4})^{3} : x^{3} + 3 x^{7} + 3 x^{11} + x^{15}

= \int x^{3} + 3 x^{7} + 3 x^{11} + x^{15} 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^{3} d x + \int 3 x^{7} d x + \int 3 x^{11} d x + \int x^{15} d x

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

\int 3 x^{7} d x = \frac{3 x ^{8}}{8}

\int 3 x^{11} d x = \frac{x ^{12}}{4}

\int x^{15} d x = \frac{x ^{16}}{16}

= \frac{x ^{4}}{4} + \frac{3 x ^{8}}{8} + \frac{x ^{12}}{4} + \frac{x ^{16}}{16}

Add a constant to the solution

= \frac{x ^{4}}{4} + \frac{3 x ^{8}}{8} + \frac{x ^{12}}{4} + \frac{x ^{16}}{16} + C

\frac{d}{d x} (sin (x + 2))

\int_{- \infty}^{0} \frac{1}{4 - 3 x} d x

\frac{s}{( s - 3 ) ^{3}}

\frac{\partial}{\partial y} (ln (2 x^{2} + 3 y^{2}))

\frac{d}{d x} (\frac{A}{x ^{9}} + B e^{x})

What is the integral of x^3(1+x^4)^3 ?
The integral of x^3(1+x^4)^3 is (x^4)/4+(3x^8)/8+(x^{12})/4+(x^{16})/(16)+C