What is the integral of 1/(2x^2+8x+20) ?

The integral of 1/(2x^2+8x+20) is 1/(2sqrt(6))arctan(1/(sqrt(6))(x+2))+C

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integral of 1/(2x^2+8x+20)

\int \frac{1}{2 x ^{2} + 8 x + 20} d x

\frac{1}{2 6} arctan (\frac{1}{6} (x + 2)) + C

Solution steps

\int \frac{1}{2 x ^{2} + 8 x + 20} d x

Complete the square

2 x^{2} + 8 x + 20 : 2 (x + 2)^{2} + 12

= \int \frac{1}{2 ( x + 2 ) ^{2} + 12} d x

Apply u-substitution

: \int \frac{1}{2 u ^{2} + 12} d u

= \int \frac{1}{2 u ^{2} + 12} d u

Apply Integral Substitution

= \int \frac{1}{2 6 ( 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}{2 6} \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}{2 6} arctan (v)

Substitute back

= \frac{1}{2 6} arctan (\frac{2}{2 3} (x + 2))

Simplify

\frac{1}{2 6} arctan (\frac{2}{2 3} (x + 2)) : \frac{1}{2 6} arctan (\frac{1}{6} (x + 2))

= \frac{1}{2 6} arctan (\frac{1}{6} (x + 2))

Add a constant to the solution

= \frac{1}{2 6} arctan (\frac{1}{6} (x + 2)) + C

\int 8^{x} d x

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What is the integral of 1/(2x^2+8x+20) ?
The integral of 1/(2x^2+8x+20) is 1/(2sqrt(6))arctan(1/(sqrt(6))(x+2))+C