Answer:
The above solution is your answer .
Integration
We need to evaluate the following integral.
[tex] \displaystyle I = \int \frac{x + 1}{x(x {e}^{x} + 2)} \ dx[/tex]
Can be rewritten as,
[tex] \displaystyle \implies I= \int \frac{ {e}^{x} (x + 1)}{x {e}^{x} (x {e}^{x} + 2)} \ dx[/tex]
Now we will use method of substitution to solve it further.
[tex] \bf{Put \ xe^x = t \implies e^x(x+1) \ dx = dt}[/tex]
[tex] \displaystyle \implies I = \int \frac{dt}{t (t + 2)}[/tex]
[tex] \displaystyle \implies I = \frac{1}{2} \int \frac{(t + 2) - t}{t (t + 2)} \ dt[/tex]
[tex] \displaystyle \implies I = \frac{1}{2} \int \frac{1}{t} - \frac{1}{t + 2} \ dt[/tex]
Now we will use the following identity:
[tex] \boxed{ \int \frac{dx}{x + a} = \log | x + a | + c }[/tex]
Using this, we get:
[tex] {\displaystyle \implies I = \frac{1}{2} \bigg( \log |t| - \log |t + 2| \bigg) + c}[/tex]
[tex] \displaystyle \implies I = \frac{1}{2} \log \left | \frac{t}{t + 2} \right| + c[/tex]
Undoing the substitution, we get:
[tex] \displaystyle \implies I = \frac{1}{2} \log \left | \frac{x {e}^{x} }{x {e}^{x} + 2} \right| + c[/tex]
This is the required answer.
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Answers & Comments
Answer:
The above solution is your answer .
Verified answer
Topic:
Integration
Solution:
We need to evaluate the following integral.
[tex] \displaystyle I = \int \frac{x + 1}{x(x {e}^{x} + 2)} \ dx[/tex]
Can be rewritten as,
[tex] \displaystyle \implies I= \int \frac{ {e}^{x} (x + 1)}{x {e}^{x} (x {e}^{x} + 2)} \ dx[/tex]
Now we will use method of substitution to solve it further.
[tex] \bf{Put \ xe^x = t \implies e^x(x+1) \ dx = dt}[/tex]
[tex] \displaystyle \implies I = \int \frac{dt}{t (t + 2)}[/tex]
[tex] \displaystyle \implies I = \frac{1}{2} \int \frac{(t + 2) - t}{t (t + 2)} \ dt[/tex]
[tex] \displaystyle \implies I = \frac{1}{2} \int \frac{1}{t} - \frac{1}{t + 2} \ dt[/tex]
Now we will use the following identity:
[tex] \boxed{ \int \frac{dx}{x + a} = \log | x + a | + c }[/tex]
Using this, we get:
[tex] {\displaystyle \implies I = \frac{1}{2} \bigg( \log |t| - \log |t + 2| \bigg) + c}[/tex]
[tex] \displaystyle \implies I = \frac{1}{2} \log \left | \frac{t}{t + 2} \right| + c[/tex]
Undoing the substitution, we get:
[tex] \displaystyle \implies I = \frac{1}{2} \log \left | \frac{x {e}^{x} }{x {e}^{x} + 2} \right| + c[/tex]
This is the required answer.