What is the integral of 1/(x^2+4x+25) ?

The integral of 1/(x^2+4x+25) is 1/(sqrt(21))arctan((x+2)/(sqrt(21)))+C

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integral of 1/(x^2+4x+25)

\int \frac{1}{x ^{2} + 4 x + 25} d x

\frac{1}{21} arctan (\frac{x + 2}{21}) + C

Solution steps

\int \frac{1}{x ^{2} + 4 x + 25} d x

Complete the square

x^{2} + 4 x + 25 : (x + 2)^{2} + 21

= \int \frac{1}{( x + 2 ) ^{2} + 21} d x

Apply u-substitution

: \int \frac{1}{u ^{2} + 21} d u

= \int \frac{1}{u ^{2} + 21} d u

Apply Integral Substitution

= \int \frac{1}{21 ( 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}{21} \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}{21} arctan (v)

Substitute back

= \frac{1}{21} arctan (\frac{x + 2}{21})

Add a constant to the solution

= \frac{1}{21} arctan (\frac{x + 2}{21}) + C

\int x 4 - x d x

\int_{0}^{1} 2 y^{2} d y

ln (e^{- x} + x e^{- x})

\frac{d}{d x} (\frac{5 x}{x - 3})

\frac{y}{12000} + 0.058 y^{^{'}} = 0.0015

What is the integral of 1/(x^2+4x+25) ?
The integral of 1/(x^2+4x+25) is 1/(sqrt(21))arctan((x+2)/(sqrt(21)))+C