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integral part 2

Note by Toha Muhammad
4 years 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. Aditya Parson · 4 years ago

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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. Ivan Stošić · 4 years ago

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