What is the integral of (16x^3)/(x^4+3) ?

The integral of (16x^3)/(x^4+3) is 4ln|x^4+3|+C

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integral of (16x^3)/(x^4+3)

\int \frac{16 x ^{3}}{x ^{4} + 3} d x

4 ln x^{4} + 3 + C

Solution steps

\int \frac{16 x ^{3}}{x ^{4} + 3} d x

Take the constant out:

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

= 16 \cdot \int \frac{x ^{3}}{x ^{4} + 3} d x

Apply u-substitution

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

= 16 \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

= 16 \cdot \frac{1}{4} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 16 \cdot \frac{1}{4} ln ∣ u ∣

Substitute back

u = x^{4} + 3

= 16 \cdot \frac{1}{4} ln x^{4} + 3

Simplify

16 \cdot \frac{1}{4} ln x^{4} + 3 : 4 ln x^{4} + 3

= 4 ln x^{4} + 3

Add a constant to the solution

= 4 ln x^{4} + 3 + C

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What is the integral of (16x^3)/(x^4+3) ?
The integral of (16x^3)/(x^4+3) is 4ln|x^4+3|+C