Question

How do solve the following integral? `int(t+7)^2/t^3dt`

`int(t+7)^2/t^3dt`

Answer 1

Just expand the function and integrate. The answer is `lnt-14/t-49/(2t^2)+C`.

Explanation:

Let `f(t)=(t+7)^2/t^3`. First, let's rearrange `f(t)` into simpler terms.
`f(t)=(t^2+14t+49)/t^3`
`=1/t+14/t^2+49/t^3`
`=t^-1+14t^-2+49t^-3`

Then integrate `f(t)`. Note that `intx^ndx=1/(n+1)x^(n+1)+C` if `n!=-1`, where `C` is the integral constant. If `n=-1`, `intx^-1dx=int(1/x)dx=lnx +C` is applied.

`intf(t)dt=int(t^-1+14t^-2+49t^-3)dt`
`=intt^-1dt+14intt^-2dt+49intt^-3dt`
`=lnt+14(-t^-1)+49(-1/2t^-2)+C`
`=lnt-14/t-49/(2t^2)+C`.

Answer 2

`ln(t)-(7(4t+7))/(2t^2)+c`

Explanation:

solving `(t+7)^2`

`(t+7)^2=t^2+14t+49`

now in the integral

`int (t^2+14t+49)/t^3 dt`

separating the fractions

`int (t^2/t^3+14t/t^3+49/t^3) dt`

`int 1/t+14/t^2+49/t^3 dt`

separting

`int 1/tdt=ln(t)+c`

`int 14/t^2dt=-14/t+c`

`int49/t^3dt=-49/(2t^2)+c`

`int 1/t+14/t^2+49/t^3 dt=ln(t)-14/t-49/(2t^2)+c`

operating

`int 1/t+14/t^2+49/t^3 dt=ln(t)-(7(4t+7))/(2t^2)+c`

`int (t^2+14t+49)/t^3 dt=ln(t)-(7(4t+7))/(2t^2)+c`