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

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

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

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

ln 1 + x^{4} + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 4 \cdot \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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 4 \cdot \frac{1}{4} ln ∣ u ∣

Substitute back

u = 1 + x^{4}

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

Simplify

4 \cdot \frac{1}{4} ln 1 + x^{4} : ln 1 + x^{4}

= ln 1 + x^{4}

Add a constant to the solution

= ln 1 + x^{4} + C

\frac{\partial}{\partial x} (x^{2} + 2)

\int_{1}^{8} \frac{5 ln ( x )}{x ^{2}} d x

n = 0 \sum \infty n - 2

y^{^{''}} + 2 y^{^{'}} + 3 y = 0, y^{^{'}} (0) = 2, y (0) = 1

y^{^{'}} + 7 y = e^{3 x}

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