Unable to understand this intergration

Question

I have exams tomorrow. It would be really helpful if someone helps me. Thank You.

edit : i had tried using the UV rule of integration. but that was getting really lengthy and didn't get any definitive answer. I also tried breaking them down and doing the integration. Again, it was lengthy and the attached image had it done in just one line. So, I came here just to clear this up. Also, I forgot about the formula that the accepted answer mentioned.

Answer 1

We know that $$\int (f (x)+ f'(x))e^x dx = f (x) e^x \tag {1}$$

Here we have $f (x)=\frac {1}{1+x^2} $. Now, let us find the derivative of $f (x) $, that is, $f'(x) $. Using the quotient rule, we have, $$\frac {d}{dx}f (x) = \frac {(1+x^2)\frac {d}{dx}(1)-(1)\frac {d}{dx}(1+x^2)}{(1+x^2)^2} $$ $$=-\frac {2x}{(1+x^2)^2} $$

Now use $(1) $ and we have the result. Hope it helps.

Answer 2

By the product rule for differentiation, you have: $$\left(\frac{1}{1+x^2}e^x\right)' =\frac{1}{1+x^2}\left(e^x\right)' + \color{blue}{\left(\frac{1}{1+x^2}\right)'}e^x = \frac{1}{1+x^2}e^x + \color{blue}{\frac{-2x}{\left(1+x^2\right)^2}}e^x$$ The blue step (derivative) is explained in your notes.

Now note that the right-hand side is exactly the integrand: $$\frac{1}{1+x^2}e^x + \frac{-2x}{\left(1+x^2\right)^2}e^x = \left( \frac{1}{1+x^2} - \frac{2x}{\left(1+x^2\right)^2}\right)e^x $$ So you simply have: $$\int \left( \frac{1}{1+x^2} - \frac{2x}{\left(1+x^2\right)^2}\right)e^x \,\mbox{d}x = \int\left(\frac{1}{1+x^2}e^x\right)'\,\mbox{d}x = \frac{1}{1+x^2}e^x+C$$

Answer 3

You dont actually integrate, you work the other way around using that $$\int_{}^{} f'(x) dx=f(x)+C$$ and differentiate the right hand side. Then you use product rule for differentiation. You get the given result.