What is the integral of 1/(sqrt(t^2-2t+10)) ?

The integral of 1/(sqrt(t^2-2t+10)) is ln(1/3 |t-1+sqrt(t^2-2t+10)|)+C

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integral of 1/(sqrt(t^2-2t+10))

\int \frac{1}{t ^{2} - 2 t + 10} d t

ln (\frac{1}{3} t - 1 + t^{2} - 2 t + 10) + C

Solution steps

\int \frac{1}{t ^{2} - 2 t + 10} d t

Complete the square

t^{2} - 2 t + 10 : (t - 1)^{2} + 9

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

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

Simplify

ln tan (arctan (\frac{1}{3} (t - 1))) + sec (arctan (\frac{1}{3} (t - 1))) : ln (\frac{1}{3} t - 1 + t^{2} - 2 t + 10)

= ln (\frac{1}{3} t - 1 + t^{2} - 2 t + 10)

Add a constant to the solution

= ln (\frac{1}{3} t - 1 + t^{2} - 2 t + 10) + C

\int cos^{7} (2 x) d x

\frac{\partial}{\partial x} (\frac{2 x}{9 - tan ( x )})

\int \frac{x ^{11}}{2 x ^{6} + 2} d x

y^{^{''}} + y^{^{'}} - 6 y = 2 x

t^{2} y^{^{'}} + t y = y

What is the integral of 1/(sqrt(t^2-2t+10)) ?
The integral of 1/(sqrt(t^2-2t+10)) is ln(1/3 |t-1+sqrt(t^2-2t+10)|)+C