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`        What is the solution for ∫√(1 + tanx) dx? Please give me answer  `
7 months ago

```							put y^2=1 + tanx2ydy= sec^2xdx=(1 + tan^2x)dx=(1 + (y^2 – 1)^2)dxso that I= ∫2y^2dy/(1 + (y^2 – 1)^2)= ∫2y^2dy/(y^4 – 2y^2+2)now this last integral is superbly tricky.write 2y^2/(y^4 – 2y^2+2) as 2/(y^2 – 2+2/y^2)= (1 – √2/y^2)/(y^2 – 2+2/y^2) + (1+√2/y^2)/(y^2 – 2+2/y^2)this way we have two integrals. in the first integral, write the denominator y^2 – 2+2/y^2 as (y+√2/y)^2 – 2(1+√2) and then substitute y+√2/y= z and the integral will turn into a standard partial fraction form.in the second interal, write the denominator y^2 – 2+2/y^2 as (y – √2/y)^2 – 2+2√2 and then substitute y – √2/y= t and the integral will turn into a standard partial fraction form.
```
7 months ago
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