Dear Student, I=∫(ln x + tan-1x)dx⇒I=∫(ln x) 1 dx+∫(tan-1x) 1 dxApplying integration by parts, we getI= x ln x-∫x 1x dx+ x (tan-1x)-∫x 11+x2dxI=x ln x-x+x (tan-1x)-∫x1+x2dxPut 1+x2=t, ⇒2x dx=dt ⇒x dx = dt2So, I=x ln x-x+x (tan-1x)-12 ∫1tdt=x ln x-x+x (tan-1x)-12 logt + C⇒I=x ln x-x+x (tan-1x) -12ln 1+x2 + C Regards,

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Dear Student,

I = \int (\ln x + \tan^{- 1} x) dx \Rightarrow I = \int (\ln x) . 1 dx + \int (\tan^{- 1} x) . 1 dx Applying integration by parts, we get I = x . \ln x - \int x . \frac{1}{x} . dx + x . (\tan^{- 1} x) - \int x . \frac{1}{1 + x^{2}} dx I = x . \ln x - x + x . (\tan^{- 1} x) - \int \frac{x}{1 + x^{2}} dx Put 1 + x^{2} = t, \Rightarrow 2 x . dx = dt \Rightarrow x dx = \frac{dt}{2} So, I = x . \ln x - x + x . (\tan^{- 1} x) - \frac{1}{2} . \int \frac{1}{t} dt = x . \ln x - x + x . (\tan^{- 1} x) - \frac{1}{2} \log |t| + C \Rightarrow I = x . \ln x - x + x . (\tan^{- 1} x) - \frac{1}{2} \ln |1 + x^{2}| + C

Regards,

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