Question

integrate dx/(1+x2)√(1-x2)
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Answer 1

Answer:

I=∫1−x2(1+x2)1+x4−−−−−√dx

I=∫x−1x(x+1x)x(x+1x)2−2−−−−−−−−−−√dx

I=∫x(1−1x2)(x+1x)x(x+1x)2−2−−−−−−−−−−√dx

I=∫1−1x2(x+1x)(x+1x)2−2−−−−−−−−−−√dx

Things are getting messy! It's time for a substitution ;)

Let x+1x=t

Taking derivative of both sides,

(1−1x2)dx=dt

I=∫1tt2−2−−−−−√dt

Now take t2–2=y2

Taking derivative of both sides,

2tdt=2ydy

dt=ytdy

I=∫yt2ydy

I=∫1y2+2dy

Integrating…

I=12√tan−1(y2√)+C

I=12√tan−1(t2–2√2√)+C

I=12√tan−1((x+1x)2–2√2√)+C

That's it!

Step-by-step explanation:

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