What is the integral of ln(t^2-4t+5) ?

The integral of ln(t^2-4t+5) is ln(t^2-4t+5)(t-2)-2(-arctan(t-2)+t-2)+C

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integral of ln(t^2-4t+5)

\int ln (t^{2} - 4 t + 5) d t

ln (t^{2} - 4 t + 5) (t - 2) - 2 (- arctan (t - 2) + t - 2) + C

Solution steps

\int ln (t^{2} - 4 t + 5) d t

Complete the square

t^{2} - 4 t + 5 : (t - 2)^{2} + 1

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

Apply u-substitution

: \int ln (u^{2} + 1) d u

= \int ln (u^{2} + 1) d u

Apply Integration By Parts

: u ln (u^{2} + 1) - \int \frac{2 u ^{2}}{u ^{2} + 1} d u

= u ln (u^{2} + 1) - \int \frac{2 u ^{2}}{u ^{2} + 1} d u

\int \frac{2 u ^{2}}{u ^{2} + 1} d u = 2 (- arctan (u) + u)

= u ln (u^{2} + 1) - 2 (- arctan (u) + u)

Substitute back

u = t - 2

= (t - 2) ln ((t - 2)^{2} + 1) - 2 (- arctan (t - 2) + t - 2)

Expand

(t - 2)^{2} + 1 : t^{2} - 4 t + 5

= ln (t^{2} - 4 t + 5) (t - 2) - 2 (- arctan (t - 2) + t - 2)

Add a constant to the solution

= ln (t^{2} - 4 t + 5) (t - 2) - 2 (- arctan (t - 2) + t - 2) + C

\frac{d}{d x} (\frac{x y}{x + y})

\int \frac{1}{sin ^{2} ( \frac{x}{2} ) cos ^{2} ( \frac{x}{2} )} d x

\frac{d}{d x} (ln (\frac{4 x}{x ^{2} + 5}))

\frac{d}{d x} (sin (x + y))

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

What is the integral of ln(t^2-4t+5) ?
The integral of ln(t^2-4t+5) is ln(t^2-4t+5)(t-2)-2(-arctan(t-2)+t-2)+C