Use intergration by parts

Evaluate the integral. (Remember to use ln |u| where appropriate.)
From [0,5]
7t/(t-6)^2 dt

∫ 7t / (t - 6)^2 dt
= 7 ∫ (t - 6 + 6) / (t - 6)^2 dt
= 7 ∫ dt / (t - 6) + 42 ∫ dt / (t - 6)^2
= 7 ln l t - 6 l - 42 / (t - 6) + c

Plugging limits, t = 0 to 5
= 7 ln 1 + 42 - (7 ln 6 + 7)
= 35 - 7 ln (6).

Integration by parts?

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I will actually do something a little different.
u=t-6
u+6=t
du=dt
7*INT[(u+6)/u^2 du]
7*INT[1/u +6u^-2 du]
7[lnu-6/u]
now you need to change your limits.
at t=0
u=-6
at t=5 u=-1
plug in -1 first
7[ln(-1)-6/-1]=0+42 Remember there is an absolute value sign.
plug in -6.
7[ln(-6)-6/-6]=7ln6+7.
42-(7ln6+7)=35-7ln6.

From [0,5]
7t/(t-6)^2 dt

Using integration by parts:

U= 7t; dV= (t-6)^-2dt
Du = 7dt; V= -(t-6)^-1

UV- INT [VDU] = -7t/(t-6) - INT [ -7/(t-6) ]dt

= -7t/(t-6) + 7 ln| t-6| | (0,5)

= [ -35/-1 + 7*ln(1)] -[ 0 + 7*ln(6)]

= 35 -7* ln (6)

= approx 22.46

Hoping this helps!

-7 (-5 + Log[6])

22.4577