What is the integral of 1/(x^2+4x+29) ?

The integral of 1/(x^2+4x+29) is 1/5 arctan((x+2)/5)+C

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integral of 1/(x^2+4x+29)

\int \frac{1}{x ^{2} + 4 x + 29} d x

\frac{1}{5} arctan (\frac{x + 2}{5}) + C

Solution steps

\int \frac{1}{x ^{2} + 4 x + 29} d x

Complete the square

x^{2} + 4 x + 29 : (x + 2)^{2} + 25

= \int \frac{1}{( x + 2 ) ^{2} + 25} d x

Apply u-substitution

: \int \frac{1}{u ^{2} + 25} d u

= \int \frac{1}{u ^{2} + 25} d u

Apply Integral Substitution

= \int \frac{1}{5 ( 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}{5} \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}{5} arctan (v)

Substitute back

= \frac{1}{5} arctan (\frac{x + 2}{5})

Add a constant to the solution

= \frac{1}{5} arctan (\frac{x + 2}{5}) + C

x^{\frac{1}{3}}

\frac{d}{d x} (- e^{x} (x + 6))

x \to 3 + lim (\frac{4 - x}{9 - x ^{2}})

\int \frac{sin ( 3 x )}{3 + 3 cos ( 3 x )} d x

\frac{d y}{d x}, 3 x y = 6

What is the integral of 1/(x^2+4x+29) ?
The integral of 1/(x^2+4x+29) is 1/5 arctan((x+2)/5)+C