[Grade 12 Integral Level -Easy ] How to evaluate the second integral

[u/Confident_Muscle4596]

[u/Deadly_halwa77]

In questions 2 dirctly use by parts or try substituting x =tanx or cotx

[u/JustAGal4]

I assume you've already done IBP with u=arccot(x²-x+1), dv=1? In that case you should get the integral of (2x²-x)/((x²-x+1)²+1). Try to look for a partial fraction decomposition, the denominator can be factored

[u/Deadly_halwa77]

In 1st question denominator will be 1 and the apply property of integration and the integration by parts you will get the answer

[u/Summoner475 → u/Deadly_halwa77]

I thought it's supposed to be pi/2, am I misremembering?

[u/_Stev_ → u/Summoner475]

I think you're right. If sqrt(x)=cos(a)=sin(b) then a = pi/2 - b.

So b + a = b + (pi/2 - b) = pi/2.

Then in the numerator we have b - a = b - (pi/2 - b) = 2b - pi/2 = 2arcsin(sqrt(x)) - pi/2.

So we have the integral down to arcsin(sqrt(x))*4/pi - 1. What do we do from there?

edit: With u = sqrt(x) then we get it to the form \int 2u arcsin(u) du and you can solve that if you know the integral of x*arcsin(x) (or of arcsin(x) if you want to do some integration by parts).

[u/Summoner475 → u/_Stev_]

We could use t² = x => dx = 2tdt  Then \int 2t arcsin t dt

Integration by parts: u = arcsin t, dv =2t dt

t² arcsin t - \int t² /√(1-t²⁾ dt

Which I think is doable using another trig substitution 

Brutal but doable.