What is the integral of 1/(sqrt(x^2+8x+19)) ?

The integral of 1/(sqrt(x^2+8x+19)) is ln| 1/(sqrt(3))(x+4)+sqrt(1/3 (x^2+8x+19))|+C

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integral of 1/(sqrt(x^2+8x+19))

\int \frac{1}{x ^{2} + 8 x + 19} d x

ln \frac{1}{3} (x + 4) + \frac{1}{3} (x^{2} + 8 x + 19) + C

Solution steps

\int \frac{1}{x ^{2} + 8 x + 19} d x

Complete the square

x^{2} + 8 x + 19 : (x + 4)^{2} + 3

= \int \frac{1}{( x + 4 ) ^{2} + 3} d x

Apply u-substitution

: \int \frac{1}{u ^{2} + 3} d u

= \int \frac{1}{u ^{2} + 3} d u

Apply Trigonometric Substitution

: \int sec (v) d v

= \int sec (v) d v

Use the common integral:

\int sec (v) d v = ln ∣ tan (v) + sec (v) ∣

= ln ∣ tan (v) + sec (v) ∣

Substitute back

= ln tan (arctan (\frac{1}{3} (x + 4))) + sec (arctan (\frac{1}{3} (x + 4)))

Simplify

ln tan (arctan (\frac{1}{3} (x + 4))) + sec (arctan (\frac{1}{3} (x + 4))) : ln \frac{1}{3} (x + 4) + \frac{1}{3} (x^{2} + 8 x + 19)

= ln \frac{1}{3} (x + 4) + \frac{1}{3} (x^{2} + 8 x + 19)

Add a constant to the solution

= ln \frac{1}{3} (x + 4) + \frac{1}{3} (x^{2} + 8 x + 19) + C

\frac{\partial}{\partial x} (ln (y x))

\int sin (r^{2}) r d r

\int \frac{6 ^{x} - 4 \cdot 1 5 ^{x}}{3 ^{2 x}} d x

ln (ln (6 s))

\frac{d y}{d x} = 3 e^{x - y}

What is the integral of 1/(sqrt(x^2+8x+19)) ?
The integral of 1/(sqrt(x^2+8x+19)) is ln| 1/(sqrt(3))(x+4)+sqrt(1/3 (x^2+8x+19))|+C