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

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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

\int cos (n \cdot x) d x

n = 1 \sum \infty \frac{1}{( 2 n + 1 ) ^{3}}

\int \frac{1}{x ^{4} + 4 x ^{2} + 4} d x

ln (w (h - l) + π - T) + ln (l)

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