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

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

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2 Answers
yumean1119
Dec 4, 2017

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#.

George M.
Dec 4, 2017

#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#

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