# integral part 2

5 years, 1 month ago

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The second one is much easier:

Substitute: $$\sqrt{1 - e^x } =z$$

Your integral will become: $$\int \frac{2z^2}{z^2 -1}$$.

Now, integrate this using partial fractions.

- 5 years, 1 month ago

For the first one:

$$\frac{3 x^4+x^3+20 x^2+3 x+31}{(x+1) \left(x^2+4\right)^2}=\frac{x}{x^2+4}-\frac{1}{\left(x^2+4\right)^2}+\frac{2}{x+1}$$

For the first part, use $$t = x^2$$. For the third part, use $$u = x+1$$. The second part is a bit tricky. Use $$x = 2 \tan{z}$$ and you will arrive at $$\frac{1}{(x^2+4)^2} dx = \frac{\cos ^2(z)}{8} dz$$, which is equal to $$\frac{1}{16} (\cos (2 z)+1) dz$$ and should now be easy to integrate.

- 5 years, 1 month ago