What is the integral of 1/(x^2+6x+25) ?

The integral of 1/(x^2+6x+25) is 1/4 arctan((x+3)/4)+C

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integral of 1/(x^2+6x+25)

\int \frac{1}{x ^{2} + 6 x + 25} d x

\frac{1}{4} arctan (\frac{x + 3}{4}) + C

Solution steps

\int \frac{1}{x ^{2} + 6 x + 25} d x

Complete the square

x^{2} + 6 x + 25 : (x + 3)^{2} + 16

= \int \frac{1}{( x + 3 ) ^{2} + 16} d x

Apply u-substitution

: \int \frac{1}{u ^{2} + 16} d u

= \int \frac{1}{u ^{2} + 16} d u

Apply Integral Substitution

= \int \frac{1}{4 ( 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}{4} \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}{4} arctan (v)

Substitute back

= \frac{1}{4} arctan (\frac{x + 3}{4})

Add a constant to the solution

= \frac{1}{4} arctan (\frac{x + 3}{4}) + C

\int \frac{9 x ^{2} - 6}{x ^{2}} d x

\frac{\partial}{\partial x} (\frac{- 8 x + 5}{2 y ^{2} + 1})

y^{^{'}} + \frac{1}{x} y = 7 x^{2}

\frac{d}{d x} (ln (x) - \frac{1}{x})

x = 0 \sum \infty sin (x)

What is the integral of 1/(x^2+6x+25) ?
The integral of 1/(x^2+6x+25) is 1/4 arctan((x+3)/4)+C