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

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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:

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

How do you find the integral of 1/(x^2sqrt(25-x^2))dx1/(x^2sqrt(25-x^2))dx?