# Evaluate : ∫(x−3)√(x2+3x−18x) dx - Mathematics

Evaluate : int(x-3)sqrt(x^2+3x-18)  dx

#### Solution

Let I= int(x-3)sqrt(x^2+3x-18x)dx

Put sqrt(x^2+3x−18)=t ⇒(x^2+3x−18) =t^2

On differentiating with respect to x, we get:

2x+3=2t(dt/dx)

x+3/2=t(dt/dx)

x+3/2+3−3=t(dt/dx)

x−3+9/2=t(dt/dx)..............(1)

The given integral can be rewritten as follows:

I=int(x−3+9/2-9/2)sqrt(x^2+3x-18)dx

=int(x-3+9/2)sqrt(x^2+3x+18)dx-9/2intsqrt(x^2+3x+18)dx..............(2)

Suppose that l_1=int(x-3+9/2)sqrt(x^2_3x-18)dx

"On using equation  "(1), we getl_1=intt^2dt=t^3/3+C_1=(x^2+3x-18)^(3/2)/3+C_1

Suppose that l_2=intsqrt(x^2+3x-18)dx

intsqrt(x^2+3x-18)dx=intsqrt((x+3/2)^2-(9/2)^2)dx

=((2x+3)/4) sqrt(x^2+3x-18)-81/8log|(2x+3)/2+sqrt(x^2+3x-18)|+C_2

l=(x^2+3x-18)^(3/2)/3-9/8(2x+3)sqrt(x^2+3x-18)+729/16log|(2x+3)/2+sqrt(x^2+3x-18)|+C

where C=C_1+C_2 is a constant.

Concept: Methods of Integration: Integration by Substitution
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2013-2014 (March) Delhi Set 1

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