What is the integral of 1/(sqrt(x^2+2x+65)) ?

The integral of 1/(sqrt(x^2+2x+65)) is ln(1/8 |x+1+sqrt(x^2+2x+65)|)+C

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integral of 1/(sqrt(x^2+2x+65))

\int \frac{1}{x ^{2} + 2 x + 65} d x

ln (\frac{1}{8} x + 1 + x^{2} + 2 x + 65) + C

Solution steps

\int \frac{1}{x ^{2} + 2 x + 65} d x

Complete the square

x^{2} + 2 x + 65 : (x + 1)^{2} + 64

= \int \frac{1}{( x + 1 ) ^{2} + 64} d x

Apply u-substitution

: \int \frac{1}{u ^{2} + 64} d u

= \int \frac{1}{u ^{2} + 64} 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}{8} (x + 1))) + sec (arctan (\frac{1}{8} (x + 1)))

Simplify

ln tan (arctan (\frac{1}{8} (x + 1))) + sec (arctan (\frac{1}{8} (x + 1))) : ln (\frac{1}{8} x + 1 + x^{2} + 2 x + 65)

= ln (\frac{1}{8} x + 1 + x^{2} + 2 x + 65)

Add a constant to the solution

= ln (\frac{1}{8} x + 1 + x^{2} + 2 x + 65) + C

\int 3 x^{2} (x - 2) (x + 3) d x

(1 + x^{2}) \frac{d y}{d x} - (2 x y + x^{2} + x^{4}) = 0

\frac{d x}{d t} - \frac{2}{3 t} x = 4 t x^{- 5}

\frac{d}{d x} (x^{2} (e^{x} - 1))

x \to \frac{π}{2} lim (2 sec (x))

What is the integral of 1/(sqrt(x^2+2x+65)) ?
The integral of 1/(sqrt(x^2+2x+65)) is ln(1/8 |x+1+sqrt(x^2+2x+65)|)+C