What is the integral of 1/(sqrt(t^2-10t+41)) ?

The integral of 1/(sqrt(t^2-10t+41)) is ln(1/4 |t-5+sqrt(t^2-10t+41)|)+C

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integral of 1/(sqrt(t^2-10t+41))

\int \frac{1}{t ^{2} - 10 t + 41} d t

ln (\frac{1}{4} t - 5 + t^{2} - 10 t + 41) + C

Solution steps

\int \frac{1}{t ^{2} - 10 t + 41} d t

Complete the square

t^{2} - 10 t + 41 : (t - 5)^{2} + 16

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

Simplify

ln tan (arctan (\frac{1}{4} (t - 5))) + sec (arctan (\frac{1}{4} (t - 5))) : ln (\frac{1}{4} t - 5 + t^{2} - 10 t + 41)

= ln (\frac{1}{4} t - 5 + t^{2} - 10 t + 41)

Add a constant to the solution

= ln (\frac{1}{4} t - 5 + t^{2} - 10 t + 41) + C

f (x) = x^{2} arctan (5 x)

\int \frac{x + 5}{( x + 2 ) ( x - 1 )} d x

y = 5 + 5 x^{2} - 2 x^{3}, (1, 8)

x \to \infty lim (\frac{2}{3 x})

\int 10000 (t - 6) d t

What is the integral of 1/(sqrt(t^2-10t+41)) ?
The integral of 1/(sqrt(t^2-10t+41)) is ln(1/4 |t-5+sqrt(t^2-10t+41)|)+C