What is the integral of 1/(sqrt(t^2-2t+26)) ?

The integral of 1/(sqrt(t^2-2t+26)) is ln(1/5 |t-1+sqrt(t^2-2t+26)|)+C

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integral of 1/(sqrt(t^2-2t+26))

\int \frac{1}{t ^{2} - 2 t + 26} d t

ln (\frac{1}{5} t - 1 + t^{2} - 2 t + 26) + C

Solution steps

\int \frac{1}{t ^{2} - 2 t + 26} d t

Complete the square

t^{2} - 2 t + 26 : (t - 1)^{2} + 25

= \int \frac{1}{( t - 1 ) ^{2} + 25} d t

Apply u-substitution

: \int \frac{1}{u ^{2} + 25} d u

= \int \frac{1}{u ^{2} + 25} 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}{5} (t - 1))) + sec (arctan (\frac{1}{5} (t - 1)))

Simplify

ln tan (arctan (\frac{1}{5} (t - 1))) + sec (arctan (\frac{1}{5} (t - 1))) : ln (\frac{1}{5} t - 1 + t^{2} - 2 t + 26)

= ln (\frac{1}{5} t - 1 + t^{2} - 2 t + 26)

Add a constant to the solution

= ln (\frac{1}{5} t - 1 + t^{2} - 2 t + 26) + C

y^{^{'}} + y^{2} = 0

\int_{\frac{1}{2}}^{5} 12 x ln (2 x) d x

\frac{3}{x}

x \to 1 lim (\frac{1}{x} + 1)

y^{^{''}} + 9 y = 26 e^{2 x}

What is the integral of 1/(sqrt(t^2-2t+26)) ?
The integral of 1/(sqrt(t^2-2t+26)) is ln(1/5 |t-1+sqrt(t^2-2t+26)|)+C