What is the integral of (2x+4)/(x^2+4x+1) ?

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

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integral of (2x+4)/(x^2+4x+1)

\int \frac{2 x + 4}{x ^{2} + 4 x + 1} d x

ln x^{2} + 4 x + 1 + C

Solution steps

\int \frac{2 x + 4}{x ^{2} + 4 x + 1} d x

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= ln ∣ u ∣

Substitute back

u = x^{2} + 4 x + 1

= ln x^{2} + 4 x + 1

Add a constant to the solution

= ln x^{2} + 4 x + 1 + C

2 cos^{2} (x)

\int \frac{1}{x} \cdot \frac{x ^{5}}{5} d x

\int \frac{1}{7 x + 11} d x

\int 3 x 1 - 4 x^{2} d x

\int cos (x) (sec (x) + tan (x)) d x

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