What is the integral of 1/(sqrt(x^2-8x+25)) ?

The integral of 1/(sqrt(x^2-8x+25)) is ln(1/3 |x-4+sqrt(x^2-8x+25)|)+C

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integral of 1/(sqrt(x^2-8x+25))

\int \frac{1}{x ^{2} - 8 x + 25} d x

ln (\frac{1}{3} x - 4 + x^{2} - 8 x + 25) + C

Solution steps

\int \frac{1}{x ^{2} - 8 x + 25} d x

Complete the square

x^{2} - 8 x + 25 : (x - 4)^{2} + 9

= \int \frac{1}{( x - 4 ) ^{2} + 9} d x

Apply u-substitution

: \int \frac{1}{u ^{2} + 9} d u

= \int \frac{1}{u ^{2} + 9} 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}{3} (x - 4))) + sec (arctan (\frac{1}{3} (x - 4)))

Simplify

ln tan (arctan (\frac{1}{3} (x - 4))) + sec (arctan (\frac{1}{3} (x - 4))) : ln (\frac{1}{3} x - 4 + x^{2} - 8 x + 25)

= ln (\frac{1}{3} x - 4 + x^{2} - 8 x + 25)

Add a constant to the solution

= ln (\frac{1}{3} x - 4 + x^{2} - 8 x + 25) + C

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

\int sin (ln (12 x)) d x

6 - x^{2}, 26 - 9 x

\frac{d}{d t} (2 sin (3 t))

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

What is the integral of 1/(sqrt(x^2-8x+25)) ?
The integral of 1/(sqrt(x^2-8x+25)) is ln(1/3 |x-4+sqrt(x^2-8x+25)|)+C