integral part 2

#HelpMe! #MathProblem #Math

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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 - 7 years, 10 months ago

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 - 7 years, 10 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$