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Question 223 of 255

Q.Evaluate: ∫1+log⁡xx(2+log⁡x)(3+log⁡x) dx\displaystyle\int \dfrac{1 + \log x}{x(2 + \log x)(3 + \log x)}\,dx

Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2016Subjective· 3mImportance★★★★★
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Substitute t=log⁡xt=\log x (so dt=dx/xdt=dx/x), then use partial fractions on the resulting rational function in tt.

I=∫1+log⁡xx(2+log⁡x)(3+log⁡x) dxI=\int\frac{1+\log x}{x(2+\log x)(3+\log x)}\,dx

Let t=log⁡xt=\log x (natural log), so dt=dxxdt=\dfrac{dx}{x}:

I=∫1+t(2+t)(3+t) dtI=\int\frac{1+t}{(2+t)(3+t)}\,dt

Partial fractions:

1+t(2+t)(3+t)=A2+t+B3+t\frac{1+t}{(2+t)(3+t)}=\frac{A}{2+t}+\frac{B}{3+t}

1+t=A(3+t)+B(2+t)1+t=A(3+t)+B(2+t)

At t=−2t=-2: 1−2=A(1)⇒A=−11-2=A(1) \Rightarrow A=-1

At t=−3t=-3: 1−3=B(−1)⇒B=21-3=B(-1) \Rightarrow B=2

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