What is the integral of 2/(x^2+10x+29) ?

The integral of 2/(x^2+10x+29) is arctan(1/2 (x+5))+C

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integral of 2/(x^2+10x+29)

\int \frac{2}{x ^{2} + 10 x + 29} d x

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

Solution steps

\int \frac{2}{x ^{2} + 10 x + 29} d x

Take the constant out:

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

= 2 \cdot \int \frac{1}{x ^{2} + 10 x + 29} d x

Complete the square

x^{2} + 10 x + 29 : (x + 5)^{2} + 4

= 2 \cdot \int \frac{1}{( x + 5 ) ^{2} + 4} d x

Apply u-substitution

: \int \frac{1}{u ^{2} + 4} d u

= 2 \cdot \int \frac{1}{u ^{2} + 4} d u

Apply Integral Substitution

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

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

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

= 2 \cdot \frac{1}{2} arctan (v)

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

(1 + x^{2}) \frac{d y}{d x} - (1 + y^{2}) = 0

f (x) = x^{2} - x + 5, at x = 5

\frac{d}{d x} \frac{1}{1 + sin ^{2} ( x )}

\frac{d}{d x} (2 e^{- x^{2}} x - 2 e^{- x^{2}} x^{3})

f (x) = 2 x^{- \frac{3}{4}}

What is the integral of 2/(x^2+10x+29) ?
The integral of 2/(x^2+10x+29) is arctan(1/2 (x+5))+C