What is the integral of (x^2)/(4x^3+5) ?

The integral of (x^2)/(4x^3+5) is 1/12 ln|4x^3+5|+C

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integral of (x^2)/(4x^3+5)

\int \frac{x ^{2}}{4 x ^{3} + 5} d x

\frac{1}{12} ln 4 x^{3} + 5 + C

Solution steps

\int \frac{x ^{2}}{4 x ^{3} + 5} 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} + 5

= \frac{1}{12} ln 4 x^{3} + 5

Add a constant to the solution

= \frac{1}{12} ln 4 x^{3} + 5 + C

f (x) = 5 x^{3} ln (x)

y^{^{''}} + 2 y^{^{'}} + 5 y = 0, y (0) = 5, y^{^{'}} (0) = - 5

x^{2} y^{^{''}} + 3 x y^{^{'}} + 4 y = 0

\frac{\partial}{\partial z} (sin (z^{3}) cos (x y))

f (x) = \frac{- 2 x}{x ^{5} - x - 1}

What is the integral of (x^2)/(4x^3+5) ?
The integral of (x^2)/(4x^3+5) is 1/12 ln|4x^3+5|+C