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# integrate x/[(7x-10-x2)3/2] w.r.t x 9 years ago

Hi Suchita,

This is a good question where you can see, how to reduce the given integral to standard integral formats.

Now try to write x, as a differetial of the Dr (which is 7-2x)

So x = 1/2{(2x-7)+7), and then split it into two integrals.

So first integral is of the form -f'(x)/f(x)3/2, which is easily integrated by the substitution of f(x) = t.

For the second integral, we can note that the Dr can be factorised as (5-x)(x-2), and the sum of the two factors is a constant = 3.

So write 7 = 7/3*{(5-x)+(x-2)}

and then split this integral again into two more integrals.

where first integral is 1/(5-x)1/2(x-2)3/2 and the second integral is 1/(x-2)1/2(5-x)3/2.

the two integrals are integrated by the same mthod.

Method for the 1st, take x common from each bracket, and you have 1/x2*(5/x-1)1/2*(1-2/x)3/2.

So for this make the sub 1-2/x = t2, and you have an integral of the form dt/t2√(t2-a2), which is a standard integral. (Integrate by taking t2 common form the root, and making the substituion 1-a2/t2 = y2)

Follow the similar method for the second integral.

And that should give you the method of integration for the above question.

Alternate Solution

There's also an easier way to integrate the given question.

Directly factorise the denominator. Take x common from each factor.

Make the substitution 1-2/x = t2, and you have an integral of the form dt/t2(a2-t2)3/2.

This is again integrated, by taking t2 common from the bracket, and making the sub a2/t2-1 = y2, and you get an integral which is straight away integrable.

Hope that helps.

All the best,

Regards,

2 years ago
You can factorise denominator
7x-10-x²= (x-5)(2-x)
and put x=5cos²a + 2sin²a
dx=-3sin2a, x= 3cos²a+2
x-5 = -3sin²a,2-x=-3cos²a
a= (tan^-1 (x-5/2-x)^1/2) and slove it