If ∫ln⁡$$exx+1+ln$$⁡$$xx21+(xln$$⁡$$x)ln$$⁡$$e2xxdx=f(x)+C$$ Where C is arbitrary Constant, then $$eef2$$−$$1$$ is $$=$$

Question

If ∫ln⁡$$exx+1+ln$$⁡$$xx21+(xln$$⁡$$x)ln$$⁡$$e2xxdx=f(x)+C$$ Where C is arbitrary Constant, then $$eef2$$−$$1$$  is $$=$$

Infinity Learn · Accepted Answer

Let $$I=$$∫ln⁡$$exx+1+ln$$⁡$$xx21+(xln$$⁡$$x)ln$$⁡$$e2xxdx$$ Use (1)log$$⁡$$m+log$$⁡$$n=log$$⁡$$mn(2)log$$⁡$$am=mlog$$⁡$$a=$$∫ln⁡$$(e)+ln$$⁡$$xx+1+(x$$⋅ln⁡$$(x))21+(x$$⋅ln⁡$$(x))⋅ln⁡$$e2+ln$$⁡$$xxdx=$$∫$$1+(x+1)ln$$⁡(x)+x(ln$$⁡$$(x))21+(x$$⋅ln⁡$$(x))⋅(2+x$$⋅ln⁡$$x)dx=$$∫$$1+ln$$⁡$$x+xln$$⁡$$x+x(ln$$⁡$$x)21+2(xln$$⁡$$x)+(xln$$⁡$$x)2dx=$$∫$$(1+xln$$⁡$$x)(1+ln$$⁡$$x)(1+xln$$⁡$$x)2dx=$$∫$$1+ln$$⁡$$x(1+xln$$⁡$$x)dx$$ ∵∫f′$$(x)f(x)=ln$$⁡|$$f(x)|$$+c=ln$$⁡(1+xln$$⁡$$x)+C$$∴$$f(x)=ln$$⁡$$(1+xln$$⁡$$x)⇒$$f(2)=ln$$⁡|$$1+2ln$$⁡$$2$$|⇒$$f(2)=ln$$⁡|$$1+ln$$⁡$$4$$|⇒$$ef(2)=1+ln$$⁡$$4$$⇒$$ef(2)$$−$$1=ln$$⁡$$4$$⇒$$eef(2)$$−$$1=4Therefore$$, the correct answer is (1).

$If \int \frac{\ln (e x^{x + 1}) + {(\ln (x^{\sqrt{x}}))}^{2}}{1 + (x \ln x) (\ln (e^{2} x^{x}))} d x = f (x) + C Where C is arbitrary Constant, then e^{(e^{f (2)} - 1)} is =$

Remember concepts with our Masterclasses.

Ready to Test Your Skills?

free study material

If ∫ln⁡exx+1+ln⁡xx21+(xln⁡x)ln⁡e2xxdx=f(x)+C Where C is arbitrary Constant, then eef2−1 is =

Remember concepts with our Masterclasses.

Ready to Test Your Skills?

free study material

$If \int \frac{\ln (e x^{x + 1}) + {(\ln (x^{\sqrt{x}}))}^{2}}{1 + (x \ln x) (\ln (e^{2} x^{x}))} d x = f (x) + C Where C is arbitrary Constant, then e^{(e^{f (2)} - 1)} is =$