What is the integral of 1/(sqrt(x^2+2x+37)) ?

The integral of 1/(sqrt(x^2+2x+37)) is ln(1/6 |x+1+sqrt(x^2+2x+37)|)+C

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integral of 1/(sqrt(x^2+2x+37))

\int \frac{1}{x ^{2} + 2 x + 37} d x

ln (\frac{1}{6} x + 1 + x^{2} + 2 x + 37) + C

Solution steps

\int \frac{1}{x ^{2} + 2 x + 37} d x

Complete the square

x^{2} + 2 x + 37 : (x + 1)^{2} + 36

= \int \frac{1}{( x + 1 ) ^{2} + 36} d x

Apply u-substitution

: \int \frac{1}{u ^{2} + 36} d u

= \int \frac{1}{u ^{2} + 36} 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}{6} (x + 1))) + sec (arctan (\frac{1}{6} (x + 1)))

Simplify

ln tan (arctan (\frac{1}{6} (x + 1))) + sec (arctan (\frac{1}{6} (x + 1))) : ln (\frac{1}{6} x + 1 + x^{2} + 2 x + 37)

= ln (\frac{1}{6} x + 1 + x^{2} + 2 x + 37)

Add a constant to the solution

= ln (\frac{1}{6} x + 1 + x^{2} + 2 x + 37) + C

\int t e^{5 t} d t

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

\int (9 x^{2} - 5 x + 3) d x

x \to 2 lim ((x - 2)^{x - 2})

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

What is the integral of 1/(sqrt(x^2+2x+37)) ?
The integral of 1/(sqrt(x^2+2x+37)) is ln(1/6 |x+1+sqrt(x^2+2x+37)|)+C