What is the integral of (x^2+4x)/(x^3+6x^2+5) ?

The integral of (x^2+4x)/(x^3+6x^2+5) is 1/3 ln|x^3+6x^2+5|+C

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integral of (x^2+4x)/(x^3+6x^2+5)

\int \frac{x ^{2} + 4 x}{x ^{3} + 6 x ^{2} + 5} d x

\frac{1}{3} ln x^{3} + 6 x^{2} + 5 + C

Solution steps

\int \frac{x ^{2} + 4 x}{x ^{3} + 6 x ^{2} + 5} d x

Apply u-substitution

: \int \frac{1}{3 u} d u

= \int \frac{1}{3 u} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} ln ∣ u ∣

Substitute back

u = x^{3} + 6 x^{2} + 5

= \frac{1}{3} ln x^{3} + 6 x^{2} + 5

Add a constant to the solution

= \frac{1}{3} ln x^{3} + 6 x^{2} + 5 + C

\int \frac{x + 1}{x ^{3} - x} d x

\frac{\partial}{\partial L} (K (α) (α)^{α} L^{1 - α})

y^{^{'}} + 4 y = 0

n = 1 \sum \infty 8 (\frac{7}{8})^{n}

\int 18 x^{2} (6 x^{3} + 5)^{\frac{1}{2}} d x

What is the integral of (x^2+4x)/(x^3+6x^2+5) ?
The integral of (x^2+4x)/(x^3+6x^2+5) is 1/3 ln|x^3+6x^2+5|+C