What is the integral of 1/((x+4)sqrt(x^2+8x+15)) ?

The integral of 1/((x+4)sqrt(x^2+8x+15)) is arctan(sqrt(x^2+8x+15))+C

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integral of 1/((x+4)sqrt(x^2+8x+15))

\int \frac{1}{( x + 4 ) x ^{2} + 8 x + 15} d x

arctan (x^{2} + 8 x + 15) + C

Solution steps

\int \frac{1}{( x + 4 ) x ^{2} + 8 x + 15} 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 + 4)^{2} - 1)

Expand

(x + 4)^{2} - 1 : x^{2} + 8 x + 15

= arctan (x^{2} + 8 x + 15)

Add a constant to the solution

= arctan (x^{2} + 8 x + 15) + C

y = 2 x^{2}, (- 1, 2)

\int x^{2} e^{- \frac{2 x}{a}} d x

x \to 0 lim (x \cdot cos (\frac{1}{x}))

3 x^{2} - 5 x + 2

x \to - 9 - lim (\frac{1}{( x + 9 ) ^{2}})

What is the integral of 1/((x+4)sqrt(x^2+8x+15)) ?
The integral of 1/((x+4)sqrt(x^2+8x+15)) is arctan(sqrt(x^2+8x+15))+C