Question

please answer this question​

Answer 1

We have the given indefinite integral ;

${:\implies \quad \displaystyle \sf \int \dfrac{x+1}{x(xe^{x}+2)}dx}$

We will use substitution hence to solving this integral

Now , put ;

${:\implies \quad \sf e^{x}=u}$

So that :

${:\implies \quad \sf dx=\dfrac{du}{u}\quad and\quad log(u)=x}$

Now , putting the values in the integral , it can be written as ;

${:\implies \quad \displaystyle \sf \int \dfrac{log(u)+1}{log(u)\{ulog(u)+2\}}\times \dfrac{du}{u}}$

Now , we will again use substitution method for making the integral easy. So put ;

${:\implies \quad \displaystyle \sf ulog(u)+2=v}$

So that ;

${:\implies \quad \displaystyle \sf du=\dfrac{dv}{log(u)+1}\quad and\quad ulog(u)=v-2}$

Now , we have ;

${:\implies \quad \displaystyle \sf \int \dfrac{\cancel{\{log(u)+1\}}}{log(u)v}\times \dfrac{dv}{\cancel{\{log(u)+1\}}u}}$

${:\implies \quad \displaystyle \sf \int \dfrac{dv}{v\{ulog(u)\}}}$

Now , putting the value of ulog(u) = v - 2

${:\implies \quad \displaystyle \sf \int \dfrac{dv}{v(v-2)}}$

Now , using partial fraction decomposition , ths given integral can be further written as ;

${:\implies \quad \displaystyle \sf \dfrac{1}{2}\int \left(\dfrac{1}{v-2}-\dfrac{1}{v}\right)dv}$

Now ,as integrals follow distributive property. So ;

${:\implies \quad \displaystyle \sf \dfrac{1}{2}\left(\int \dfrac{1}{v-2}dv-\int \dfrac{1}{v}dv\right)}$

${:\implies \quad \displaystyle \sf \dfrac{1}{2}(log|v-2|-log|v|)+C}$

${:\implies \quad \displaystyle \sf \dfrac{1}{2}log\bigg|\dfrac{v-2}{v}\bigg| +C}$

Putting value of v ;

${:\implies \quad \displaystyle \sf \dfrac{1}{2}log\bigg|\dfrac{ulog(u)+2-2}{ulog(u)+2}\bigg| +C}$

Now, putting value of u ;

${:\implies \quad \displaystyle \sf \dfrac{1}{2}log\bigg|\dfrac{e^{x}log(e^{x})}{e^{x}log(e^{x})+2}\bigg| +C}$

${:\implies \quad \displaystyle \sf \dfrac{1}{2}log\bigg|\dfrac{xe^{x}}{xe^{x}+2}\bigg| +C}$

${:\implies \quad \therefore \displaystyle \underline{\underline{\int \bf \dfrac{x+1}{x(xe^{x}+2)}dx=\dfrac{1}{2}log\bigg|\dfrac{xe^{x}}{xe^{x}+2}\bigg| +C}}}$

Used Concepts :-

• ${\boxed{\displaystyle \bf \int \dfrac{1}{x}dx=log|x|+C}}$

• ${\boxed{\displaystyle \bf \dfrac{d}{dx}(u\cdot v)=v\dfrac{du}{dx}+u\dfrac{dv}{dx}}}$

• ${\boxed{\displaystyle \bf log(a)-log(b)=log\left(\dfrac{a}{b}\right)}}$

• ${\boxed{\displaystyle \bf log(a^b)=blog(a)}}$

• ${\boxed{\displaystyle \bf \dfrac{d}{dx}\{log(x)\}=\dfrac{1}{x}}}$

• ${\boxed{\displaystyle \bf \dfrac{d}{dx}(e^x)=e^{x}}}$