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

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

integral of (x^2+2x+1)/(x^3+3x^2+3x)

\int \frac{x ^{2} + 2 x + 1}{x ^{3} + 3 x ^{2} + 3 x} d x

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

Solution steps

\int \frac{x ^{2} + 2 x + 1}{x ^{3} + 3 x ^{2} + 3 x} d x

Apply u-substitution

= \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} + 3 x^{2} + 3 x

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

Add a constant to the solution

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

\int \frac{( 1 - x )}{( 1 + x )} d x

\int x^{3} (x^{4} - 7)^{4} d x

\int \frac{x ^{3} + x ^{2} - x + 1}{x ^{3}} d x

in t e g r a l \frac{1}{3 x}

\int \frac{2}{e ^{- t}} d t

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