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

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

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

integral

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

\frac{1}{4} ln x^{4} + 1 + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{4 u} d u

= \int \frac{1}{4 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{4} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{4} ln ∣ u ∣

Substitute back

u = x^{4} + 1

= \frac{1}{4} ln x^{4} + 1

Add a constant to the solution

= \frac{1}{4} ln x^{4} + 1 + C

e^{3 x - 1}

e^{2}

e^{3 x - 2}

f (x) = (\frac{x}{x + 40})^{\frac{1}{2}}

\frac{x}{( x ^{2} + 6 ) 4}

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