What is the integral of 1/(x^2+2x+82) ?

The integral of 1/(x^2+2x+82) is 1/9 arctan((x+1)/9)+C

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integral of 1/(x^2+2x+82)

\int \frac{1}{x ^{2} + 2 x + 82} d x

\frac{1}{9} arctan (\frac{x + 1}{9}) + C

Solution steps

\int \frac{1}{x ^{2} + 2 x + 82} d x

Complete the square

x^{2} + 2 x + 82 : (x + 1)^{2} + 81

= \int \frac{1}{( x + 1 ) ^{2} + 81} d x

Apply u-substitution

: \int \frac{1}{u ^{2} + 81} d u

= \int \frac{1}{u ^{2} + 81} d u

Apply Integral Substitution

= \int \frac{1}{9 ( 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}{9} \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}{9} arctan (v)

Substitute back

= \frac{1}{9} arctan (\frac{x + 1}{9})

Add a constant to the solution

= \frac{1}{9} arctan (\frac{x + 1}{9}) + C

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

\frac{d}{d x} ((\frac{1}{3})^{2})

y^{^{'''}} - 6 y^{^{''}} = 3 - cos (x)

\frac{\partial}{\partial y} (y x^{2} + y^{2})

f (x) = - 2 x^{- 1} tan (x)

What is the integral of 1/(x^2+2x+82) ?
The integral of 1/(x^2+2x+82) is 1/9 arctan((x+1)/9)+C