What is the integral from 3 to 5 of (2t)/(t^2-4t+4) ?

The integral from 3 to 5 of (2t)/(t^2-4t+4) is 2(ln(3)+4/3)

integral from 3 to 5 of (2t)/(t^2-4t+4)

\int_{3}^{5} \frac{2 t}{t ^{2} - 4 t + 4} d t

2 (ln (3) + \frac{4}{3})

Decimal

4.86389 \dots

Solution steps

\int_{3}^{5} \frac{2 t}{t ^{2} - 4 t + 4} d t

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int_{3}^{5} \frac{t}{t ^{2} - 4 t + 4} d t

Apply Integration By Parts

= 2 [- \frac{t}{t - 2} - \int - \frac{1}{t - 2} d t]_{3}^{5}

\int - \frac{1}{t - 2} d t = - ln ∣ t - 2 ∣

= 2 [- \frac{t}{t - 2} - (- ln ∣ t - 2 ∣)]_{3}^{5}

Simplify

= 2 [- \frac{t}{t - 2} + ln ∣ t - 2 ∣]_{3}^{5}

Compute the boundaries:

ln (3) + \frac{4}{3}

= 2 (ln (3) + \frac{4}{3})

d er i v a t i v e 2 e^{x} + \frac{2}{3 x}

y^{^{'}} = \frac{4 x ^{3} + y ^{3}}{x y ^{2}}

\int (x^{2} + x)^{2} d x

n = 0 \sum \infty \frac{1}{2 ( 2 n + 1 )}

x \to 0 lim (\frac{50 x ^{2}}{sin ( x ) + 50 x ^{2}})

What is the integral from 3 to 5 of (2t)/(t^2-4t+4) ?
The integral from 3 to 5 of (2t)/(t^2-4t+4) is 2(ln(3)+4/3)