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

The integral of (x^3)/1+x^4 is (x^4)/4+(x^5)/5+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

\int \frac{x ^{3}}{1} + x^{4} d x

\frac{x ^{4}}{4} + \frac{x ^{5}}{5} + C

Solution steps

\int \frac{x ^{3}}{1} + x^{4} d x

Simplify

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

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

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

= \frac{x ^{4}}{4} + \frac{x ^{5}}{5}

Add a constant to the solution

= \frac{x ^{4}}{4} + \frac{x ^{5}}{5} + C

What is the integral of (x^3)/1+x^4 ?
The integral of (x^3)/1+x^4 is (x^4)/4+(x^5)/5+C