What is the integral of 1/(x^2+12x+61) ?

The integral of 1/(x^2+12x+61) is 1/5 arctan((x+6)/5)+C

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integral of 1/(x^2+12x+61)

\int \frac{1}{x ^{2} + 12 x + 61} d x

\frac{1}{5} arctan (\frac{x + 6}{5}) + C

Solution steps

\int \frac{1}{x ^{2} + 12 x + 61} d x

Complete the square

x^{2} + 12 x + 61 : (x + 6)^{2} + 25

= \int \frac{1}{( x + 6 ) ^{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 + 6}{5})

Add a constant to the solution

= \frac{1}{5} arctan (\frac{x + 6}{5}) + C

x \to 6 lim (\frac{2}{x - 6})

\frac{3 d u}{d t} = u^{2}

t sin^{2} (t)

y = \frac{x ^{3}}{x + 3}, (3, \frac{9}{2})

\int \frac{2 cos ( x )}{sin ^{3} ( x )} d x

What is the integral of 1/(x^2+12x+61) ?
The integral of 1/(x^2+12x+61) is 1/5 arctan((x+6)/5)+C