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