What is the integral of 1/(sqrt(t^2-10t+29)) ?

The integral of 1/(sqrt(t^2-10t+29)) is ln(1/2 |t-5+sqrt(t^2-10t+29)|)+C

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integral of 1/(sqrt(t^2-10t+29))

\int \frac{1}{t ^{2} - 10 t + 29} d t

ln (\frac{1}{2} t - 5 + t^{2} - 10 t + 29) + C

Solution steps

\int \frac{1}{t ^{2} - 10 t + 29} d t

Complete the square

t^{2} - 10 t + 29 : (t - 5)^{2} + 4

= \int \frac{1}{( t - 5 ) ^{2} + 4} d t

Apply u-substitution

: \int \frac{1}{u ^{2} + 4} d u

= \int \frac{1}{u ^{2} + 4} 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}{2} (t - 5))) + sec (arctan (\frac{1}{2} (t - 5)))

Simplify

ln tan (arctan (\frac{1}{2} (t - 5))) + sec (arctan (\frac{1}{2} (t - 5))) : ln (\frac{1}{2} t - 5 + t^{2} - 10 t + 29)

= ln (\frac{1}{2} t - 5 + t^{2} - 10 t + 29)

Add a constant to the solution

= ln (\frac{1}{2} t - 5 + t^{2} - 10 t + 29) + C

\frac{d ^{2} y}{d x ^{2}} = (\frac{d y}{d x})^{2}

4 y^{^{''}} + 7 y = 0

\frac{\partial}{\partial x} (- 3 y)

\frac{\partial}{\partial x} (e^{2 x} - x cos (t))

5 \frac{d y}{d x} + 30 y = 6

What is the integral of 1/(sqrt(t^2-10t+29)) ?
The integral of 1/(sqrt(t^2-10t+29)) is ln(1/2 |t-5+sqrt(t^2-10t+29)|)+C