What is the integral of 1/(sqrt(x^2+6x+25)) ?

The integral of 1/(sqrt(x^2+6x+25)) is ln(1/4 |x+3+sqrt(x^2+6x+25)|)+C

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integral of 1/(sqrt(x^2+6x+25))

\int \frac{1}{x ^{2} + 6 x + 25} d x

ln (\frac{1}{4} x + 3 + x^{2} + 6 x + 25) + C

Solution steps

\int \frac{1}{x ^{2} + 6 x + 25} d x

Complete the square

x^{2} + 6 x + 25 : (x + 3)^{2} + 16

= \int \frac{1}{( x + 3 ) ^{2} + 16} d x

Apply u-substitution

: \int \frac{1}{u ^{2} + 16} d u

= \int \frac{1}{u ^{2} + 16} 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}{4} (x + 3))) + sec (arctan (\frac{1}{4} (x + 3)))

Simplify

ln tan (arctan (\frac{1}{4} (x + 3))) + sec (arctan (\frac{1}{4} (x + 3))) : ln (\frac{1}{4} x + 3 + x^{2} + 6 x + 25)

= ln (\frac{1}{4} x + 3 + x^{2} + 6 x + 25)

Add a constant to the solution

= ln (\frac{1}{4} x + 3 + x^{2} + 6 x + 25) + C

\frac{1}{x + 1}

\int sin^{2} (\frac{π x}{L}) d x

t^{2} y^{^{''}} - 3 t y^{^{'}} + 4 y = 0

\frac{\partial}{\partial y} ((x + y)^{5} + (x - y)^{5})

What is the integral of 1/(sqrt(x^2+6x+25)) ?
The integral of 1/(sqrt(x^2+6x+25)) is ln(1/4 |x+3+sqrt(x^2+6x+25)|)+C