Question
Evaluate: `int (1+logx)/(x(2+logx)(3+logx))dx`
Solution
`int(1+logx)/(x(2+logx)(3+logx))dx`
Substitute logx = t..................(1)
`therefore 1/xdx=dt`
Hence, the integral becomes
`int(1+t)/((2+t)(3+t))dt`
`=int(2+t-1)/((2+t)(3+t))dt`
`=int(2+t)/((2+t)(3+t))dt-int1/((2+t)(3+t))dt`
`=int1/(3+t)dt-int((t+3)-(t+2))/((2+t)(3+t))dt`
`=int1/(3+t)dt-[int(t+3)/((2+t)(3+t))dt-int(t+2)/((2+t)(3+t))dt]`
`=int1/(3+t)dt-int1/(2+t)dt+int1/(3+t)dt`
`=2int1/(3+t)dt-int1/(2+t)dt`
`=2int1/(3+t)dt-int1/(2+t)dt`
Substituting the value of 't' from (1), we get
`int(1+logx)/(x(2+logx)(3+logx))dx`
`2ln (3+ logx )-ln( 2+ logx)+ C`
`=log|(3+logx)^2/(2 + logx)| + C`
Is there an error in this question or solution?
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Solution Evaluate : ∫(1+logx)/(x(2+logx)(3+logx))dx Concept: Evaluation of Definite Integrals by Substitution.
