How do you find the integral of #1/(x^2sqrt(25-x^2))dx#?

Question

11167 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-03-01T16:35:34

The answer is: #-sqrt(25-x^2)/(25x)+c#.

We have make a substitution:

#x=5sintrArrdx=5costdt#,

so:

#int1/(x^2sqrt(25-x^2))dx=int1/((5sint)^2sqrt(25-25sin^2t))*5costdt=#

#=int(5cost)/((5sint)^2sqrt(25(1-sin^2t)))dt=#

#=int(5cost)/(25sin^2t*5cost)dt=1/25int1/sin^2tdt=#

#=-1/25int-1/sin^2tdt=-1/25cott+c=-1/25cost/sint+c=(1)#

since

#sint=x/5# and #cost=sqrt(1-sin^2t)=sqrt(1-x^2/25)=sqrt(25-x^2)/5#

than:

#(1)=-1/25sqrt(25-x^2)/5*5/x+c=-sqrt(25-x^2)/(25x)+c#.