Verified answer

First find the indefinite integral

∫ 1/ t(t-2) dt

Split up into partial fractions

= ∫ 1/ 2(t-2) - 1/(2t) dt

= ½ ln(t-2) - ½ ln(t)

= ½ ( ln(t-2) - ln(t) )

= ½ ln( (t-2)/t )

To find the definite integral:

= ½ [ ln ( (t-2)/t ) ] ...limits {3, ∞}

= ½ ( limit as s->∞ of ln[ (s-2)/s ] ) - ln(1/3)

= ½ ( 0 - ln(1/3) )

= -½ ( ln(1) - ln(3) )

= -½ ( -ln(3) )

= ½ ln(3)

Computed in Mathematica:

Integrate[1/(t^2 - 2 t), {t, 3, Infinity}]

Log[3]/2

So, the answer would be b.

If you want to do it by hand, you should set α for the upper bound of the integral. And the indefinite integral of 1/(t^2 - 2t) is

1/2 * Log[x-2] - Log[x] -- Upper boundary: α, Lower boundary: 3

That would be:

1/2 * (Log[α-2] - Log[α] + Log[3])

Then take the limit of this expression as α approaches Infinity.

b. 1/2 ln3

? e²dt / (e²t - 4e^t + 3) e²? dt / (e²t - 4e^t + 3) This essential can't be computed making use of prevalent mathematical applications, which potential you have probably made a mistake in typing the subject.

∫[3, inf] dt / (t^2 - 2t)

Take 1/(t^2 - 2t) = 1/(t^2 - 2t + 1 - 1) = 1/((t - 1)^2 - 1)

let x = t - 1, dx = dt

= ∫[3, inf] dx / (x^2 - 1)

= ∫[3, inf] (0.5/(x - 1) - 0.5/(x + 1)) dx

= |0.5 ln(x - 1) - 0.5 ln(x + 1)| [3, ifn]

= |0.5 ln(t - 2) - 0.5 ln(t)| [3, inf]

= (0.5 ln(inf - 2) - 0.5 ln(inf)) - (0.5 ln(3 - 2) - 0.5 ln(3))

= 0 - (0 - 0.5 ln(3))

= 0.5 ln(3). (option b).