What is the integral of (2x^3+3x)/(x^4+3x^2+7) ?

The integral of (2x^3+3x)/(x^4+3x^2+7) is 1/2 ln|x^4+3x^2+7|+C

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integral of (2x^3+3x)/(x^4+3x^2+7)

\int \frac{2 x ^{3} + 3 x}{x ^{4} + 3 x ^{2} + 7} d x

\frac{1}{2} ln x^{4} + 3 x^{2} + 7 + C

Solution steps

\int \frac{2 x ^{3} + 3 x}{x ^{4} + 3 x ^{2} + 7} d x

Apply u-substitution

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

= \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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} ln ∣ u ∣

Substitute back

u = x^{4} + 3 x^{2} + 7

= \frac{1}{2} ln x^{4} + 3 x^{2} + 7

Add a constant to the solution

= \frac{1}{2} ln x^{4} + 3 x^{2} + 7 + C

\int \frac{31}{x ^{3} - 125} d x

\frac{d}{d x} (\frac{3 x - 3 sin ( x )}{- e ^{x}})

\int (1 + x^{3})^{- 1} d x

\frac{e ^{- a z}}{z + 2}

What is the integral of (2x^3+3x)/(x^4+3x^2+7) ?
The integral of (2x^3+3x)/(x^4+3x^2+7) is 1/2 ln|x^4+3x^2+7|+C