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

The integral of 1/(x^2+2x+4) is 1/(sqrt(3))arctan((x+1)/(sqrt(3)))+C

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

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

\frac{1}{3} arctan (\frac{x + 1}{3}) + C

Solution steps

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

Complete the square

x^{2} + 2 x + 4 : (x + 1)^{2} + 3

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

Apply u-substitution

: \int \frac{1}{u ^{2} + 3} d u

= \int \frac{1}{u ^{2} + 3} d u

Apply Integral Substitution

= \int \frac{1}{3 ( v ^{2} + 1 )} d v

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}{v ^{2} + 1} d v

Use the common integral:

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

= \frac{1}{3} arctan (v)

Substitute back

= \frac{1}{3} arctan (\frac{x + 1}{3})

Add a constant to the solution

= \frac{1}{3} arctan (\frac{x + 1}{3}) + C

\int cos (x) cos (\frac{x}{2}) d x

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

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\int \frac{x + 5}{x ^{2} - 6 x + 13} d x

\int \frac{3 x ^{4} + 3 x ^{2} + 1}{x ^{2} + 1} d x

What is the integral of 1/(x^2+2x+4) ?
The integral of 1/(x^2+2x+4) is 1/(sqrt(3))arctan((x+1)/(sqrt(3)))+C