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

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

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

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

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

Solution steps

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

Complete the square

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

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

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} (t - 3))) + sec (arctan (\frac{1}{4} (t - 3)))

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (cos (ln (3 x)))

\int x - x d x

2 - x^{2}, - x

x \to - 1 lim (4 x^{2})

\frac{\partial}{\partial x} (4 x^{2} e^{a y} + 3 e^{b x})

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