Solve the integral. ∫1+ln⁡xx+xln2xdx

Question

Solve the integral.   ∫1+ln⁡xx+xln2xdx

Ben Owens · Accepted Answer

Step 1  The given integral is  I=∫1+ln⁡(x)x+xln2xdx  The integral can be calculated by doing substitutions.  Step 2  Solving the integral   I=∫1+ln⁡(x)x+xln2xdx  Let u=ln⁡x⇒du=1xdx  I=∫1+u1+u2du  =∫11+u2du+∫u1+u2du  =tan−1(u)+∫u1+u2du+C1   (∵∫1x2+a2dx=1atan−1(xa))  Now let 1+u2=v⇒2udu=dv  ⇒I=tan−1(u)+12∫1vdv+C1  =tan−1(u)+12ln⁡(v)+C1+C2  =tan−1(u)+12ln⁡(v)+C   (let C1+C2=C)  Where C1 and C2 are integration constants.  Step 3  Now substituting back the values of u and v to get the required integral.  I=tan−1(u)+12ln⁡(v)+C  =tan−1(ln⁡(x))+12ln⁡(1+u2)+C  =tan−1(ln⁡(x))+12ln⁡(1+ln2(x))+C

Elaine Verrett · Accepted Answer

∫ln⁡(x)+1xln2(x)+xdx  =∫u+1u2+1du  =∫(uu2+1+1u2+1)du  =∫uu2+1du+∫1u2+1du  ∫uu2+1du  =12∫1vdv  ∫1vdv  =ln⁡(v)  12∫1vdv  =ln⁡(v)2  =ln⁡(u2+1)2  ∫1u2+1du  =arctan⁡(u)  ∫uu2+1du+∫1u2+1du  =ln⁡(u2+1)2+arctan⁡(u)  =ln⁡(ln2(x)+1)2+arctan⁡(ln⁡(x))  ∫ln⁡(x)+1xln2(x)+xdx  ln⁡(ln2(x)+1)2+arctan⁡(ln⁡(x))+C

Solve the integral. \int \frac{1+\ln x}{x+x\ln^{2}x}dx

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