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

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

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

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

ln x^{4} + 1 + C

Solution steps

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

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

Simplify

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

= ln x^{4} + 1

Add a constant to the solution

= ln x^{4} + 1 + C

\frac{( 2 t + 1 ) ^{\frac{3}{2}}}{3}

\frac{d y}{d x} = 2^{x} ln (2) + \frac{2}{x ^{2} + 1}

\frac{d}{d x} (6 (5 x^{2} + 2) x^{4} + 5)

x^{2} + 4, - x^{2} - 1, - 2, 3

\int x^{4} e^{- x^{5}} d x

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