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

The integral of (x^2)/(4x^3+7) is 1/12 ln|4x^3+7|+C

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

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

\frac{1}{12} ln 4 x^{3} + 7 + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{12 u} d u

= \int \frac{1}{12 u} d u

Take the constant out:

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

= \frac{1}{12} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{12} ln ∣ u ∣

Substitute back

u = 4 x^{3} + 7

= \frac{1}{12} ln 4 x^{3} + 7

Add a constant to the solution

= \frac{1}{12} ln 4 x^{3} + 7 + C

\frac{d}{d x} (\frac{1}{3} tan^{3} (\frac{1}{x}))

(x^{2} - x) y^{^{'}} = y - 2 x y

cos^{4} (θ + 5)

n = 1 \sum \infty \frac{n}{n ^{3} + 1}

\frac{2}{3 x + 1}

What is the integral of (x^2)/(4x^3+7) ?
The integral of (x^2)/(4x^3+7) is 1/12 ln|4x^3+7|+C