What is the integral of (8(2x^3+3x))/(x^4+3x^2-1) ?

The integral of (8(2x^3+3x))/(x^4+3x^2-1) is 4ln|x^4+3x^2-1|+C

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integral of (8(2x^3+3x))/(x^4+3x^2-1)

\int \frac{8 ( 2 x ^{3} + 3 x )}{x ^{4} + 3 x ^{2} - 1} d x

4 ln x^{4} + 3 x^{2} - 1 + C

Solution steps

\int \frac{8 ( 2 x ^{3} + 3 x )}{x ^{4} + 3 x ^{2} - 1} d x

Take the constant out:

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

= 8 \cdot \int \frac{2 x ^{3} + 3 x}{x ^{4} + 3 x ^{2} - 1} d x

Apply u-substitution

: \int \frac{1}{2 u} d u

= 8 \cdot \int \frac{1}{2 u} d u

Take the constant out:

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

= 8 \cdot \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 8 \cdot \frac{1}{2} ln ∣ u ∣

Substitute back

u = x^{4} + 3 x^{2} - 1

= 8 \cdot \frac{1}{2} ln x^{4} + 3 x^{2} - 1

Simplify

8 \cdot \frac{1}{2} ln x^{4} + 3 x^{2} - 1 : 4 ln x^{4} + 3 x^{2} - 1

= 4 ln x^{4} + 3 x^{2} - 1

Add a constant to the solution

= 4 ln x^{4} + 3 x^{2} - 1 + C

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What is the integral of (8(2x^3+3x))/(x^4+3x^2-1) ?
The integral of (8(2x^3+3x))/(x^4+3x^2-1) is 4ln|x^4+3x^2-1|+C