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Question: Evaluate the given integral: \(\int{\dfrac{\left( x+1 \right)dx}{x\left( 1+x{{e}^{x}} \right)}}\)...

Evaluate the given integral: ∫(x+1)dxx(1+xex)\int{\dfrac{\left( x+1 \right)dx}{x\left( 1+x{{e}^{x}} \right)}}

Explanation

Solution

Hint: Start by simplification of the integral by multiplying the numerator and the denominator by ex{{e}^{x}}. After simplification, let the ex{{e}^{x}} be t and solve the integral.

Complete step-by-step solution -
Before starting the solution, let us discuss the important formulas of indefinite integration.
Some important formulas are:
∫exdx=ex+c\int{{{e}^{x}}dx}={{e}^{x}}+c
∫xndx=xn+1n+1\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}}
Now let us start with the integral given in the above question.
∫(x+1)dxx(1+xex)\int{\dfrac{\left( x+1 \right)dx}{x\left( 1+x{{e}^{x}} \right)}}
Now we will divide and multiply the above integral by ex{{e}^{x}} . On doing so, we get
∫ex(x+1)dxxex(1+xex)\int{\dfrac{{{e}^{x}}\left( x+1 \right)dx}{x{{e}^{x}}\left( 1+x{{e}^{x}} \right)}}
Now to convert the integral to a form from where we can directly integrate it, we let xexx{{e}^{x}} to be t.
∴d(xex)dx=dtdx\therefore \dfrac{d\left( x{{e}^{x}} \right)}{dx}=\dfrac{dt}{dx}
⇒ex(1+x)dx=dt\Rightarrow {{e}^{x}}\left( 1+x \right)dx=dt
So, if we substitute the terms in the integral in terms of t, our integral becomes:
∫dtt(1+t)\int{\dfrac{dt}{t\left( 1+t \right)}}
Now adding and subtracting t to the numerator of the above integral, we get
∫(1+t−t)dtt(1+t)\int{\dfrac{\left( 1+t-t \right)dt}{t\left( 1+t \right)}}
=∫((1+t)t(1+t)−tt(1+t))dt=\int{\left( \dfrac{\left( 1+t \right)}{t\left( 1+t \right)}-\dfrac{t}{t\left( 1+t \right)} \right)dt}
Now we know that ∫(f(x)−g(x))dx=∫f(x)dx−∫g(x)dx\int{\left( f\left( x \right)-g\left( x \right) \right)dx}=\int{f\left( x \right)dx-\int{g\left( x \right)dx}} .
∫(1+t)t(1+t)dt−∫tt(1+t)dt\int{\dfrac{\left( 1+t \right)}{t\left( 1+t \right)}dt}-\int{\dfrac{t}{t\left( 1+t \right)}dt}
=∫1tdt−∫1(1+t)dt=\int{\dfrac{1}{t}dt}-\int{\dfrac{1}{\left( 1+t \right)}dt}
Now we know that ∫dxx\int{\dfrac{dx}{x}} is equal to lnx +c. So, our integral comes out to be:
ln⁡(t)−ln⁡(1+t)+c\ln \left( t \right)-\ln \left( 1+t \right)+c
Now we will substitute the value of t as assumed by us to convert our answer in terms of x.
lnxe⁡x−ln⁡(1+xex)+c{{\operatorname{lnxe}}^{x}}-\ln \left( 1+x{{e}^{x}} \right)+c
Now we know that ln⁡a−ln⁡b=ln⁡ab\ln a-\ln b=\ln \dfrac{a}{b} .
ln⁡xex1+xex+c\ln \dfrac{x{{e}^{x}}}{1+x{{e}^{x}}}+c
So, we can say that the value of ∫(x+1)dxx(1+xex)\int{\dfrac{\left( x+1 \right)dx}{x\left( 1+x{{e}^{x}} \right)}} is equal to ln⁡xex1+xex+c\ln \dfrac{x{{e}^{x}}}{1+x{{e}^{x}}}+c .

Note: Don’t forget to substitute the assumed variable in your integrated expression to reach the final answer. Also, you need to remember all the basic formulas that we use for indefinite integrals as they are used in definite integrations as well.