What is the integral of 1/(sqrt(x^2+14x)) ?

The integral of 1/(sqrt(x^2+14x)) is ln(1/7 |sqrt(x^2+14x)+x+7|)+C

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integral of 1/(sqrt(x^2+14x))

\int \frac{1}{x ^{2} + 14 x} d x

ln (\frac{1}{7} x^{2} + 14 x + x + 7) + C

Solution steps

\int \frac{1}{x ^{2} + 14 x} d x

Complete the square

x^{2} + 14 x : (x + 7)^{2} - 49

= \int \frac{1}{( x + 7 ) ^{2} - 49} d x

Apply u-substitution

: \int \frac{1}{u ^{2} - 49} d u

= \int \frac{1}{u ^{2} - 49} 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}{7} (x + 7))) + sec (arcsec (\frac{1}{7} (x + 7)))

Simplify

ln tan (arcsec (\frac{1}{7} (x + 7))) + sec (arcsec (\frac{1}{7} (x + 7))) : ln (\frac{1}{7} x^{2} + 14 x + x + 7)

= ln (\frac{1}{7} x^{2} + 14 x + x + 7)

Add a constant to the solution

= ln (\frac{1}{7} x^{2} + 14 x + x + 7) + C

\int e^{2 u} d u

\int (2 t - 3 t^{2}) 4 + 36 t^{2} d t

\int x \cdot ln (x^{2}) d x

x^{2}, 6 x - 9, y = 0

y^{^{'}} - \frac{2}{x} y = \frac{y ^{3}}{x ^{3}}, y (1) = 1

What is the integral of 1/(sqrt(x^2+14x)) ?
The integral of 1/(sqrt(x^2+14x)) is ln(1/7 |sqrt(x^2+14x)+x+7|)+C