What is the integral of 1/((x+3)sqrt(x^2+6x+8)) ?

The integral of 1/((x+3)sqrt(x^2+6x+8)) is arctan(sqrt(x^2+6x+8))+C

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integral of 1/((x+3)sqrt(x^2+6x+8))

\int \frac{1}{( x + 3 ) x ^{2} + 6 x + 8} d x

arctan (x^{2} + 6 x + 8) + C

Solution steps

\int \frac{1}{( x + 3 ) x ^{2} + 6 x + 8} d x

Apply u-substitution

: \int \frac{1}{u u ^{2} - 1} d u

= \int \frac{1}{u u ^{2} - 1} d u

Apply u-substitution

: \int \frac{1}{v ^{2} + 1} d v

= \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

= arctan (v)

Substitute back

= arctan ((x + 3)^{2} - 1)

Expand

(x + 3)^{2} - 1 : x^{2} + 6 x + 8

= arctan (x^{2} + 6 x + 8)

Add a constant to the solution

= arctan (x^{2} + 6 x + 8) + C

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(f (4 x))^{2} + 5

\frac{d}{d x} (lo g_{12} (x))

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

What is the integral of 1/((x+3)sqrt(x^2+6x+8)) ?
The integral of 1/((x+3)sqrt(x^2+6x+8)) is arctan(sqrt(x^2+6x+8))+C