What is the integral of 1/(sqrt(x^2+16x)) ?

The integral of 1/(sqrt(x^2+16x)) is ln(1/8 |sqrt(x^2+16x)+x+8|)+C

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integral of 1/(sqrt(x^2+16x))

\int \frac{1}{x ^{2} + 16 x} d x

ln (\frac{1}{8} x^{2} + 16 x + x + 8) + C

Solution steps

\int \frac{1}{x ^{2} + 16 x} d x

Complete the square

x^{2} + 16 x : (x + 8)^{2} - 64

= \int \frac{1}{( x + 8 ) ^{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 (arcsec (\frac{1}{8} (x + 8))) + sec (arcsec (\frac{1}{8} (x + 8)))

Simplify

ln tan (arcsec (\frac{1}{8} (x + 8))) + sec (arcsec (\frac{1}{8} (x + 8))) : ln (\frac{1}{8} x^{2} + 16 x + x + 8)

= ln (\frac{1}{8} x^{2} + 16 x + x + 8)

Add a constant to the solution

= ln (\frac{1}{8} x^{2} + 16 x + x + 8) + C

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

x \to 5 lim (\frac{6}{x})

\int \frac{1}{1 + tan ( x )} d x

y^{^{''}} + 6 y^{^{'}} = 0, y (0) = 1, y^{^{'}} (0) = 1

\int x \cdot e^{- 3 x} d x

What is the integral of 1/(sqrt(x^2+16x)) ?
The integral of 1/(sqrt(x^2+16x)) is ln(1/8 |sqrt(x^2+16x)+x+8|)+C