If ∫0π4tanxdx=aπ+bln(2+1) , then a+b=

∫0π4tanxdx,tanx=t2 =∫012t2t4+1dt =∫01t2+1+t2−1t4+1dt =∫011+1t2t2+1t2dt+∫011−1t2t2+1t2dt =∫01d(t−1t)(t−1t)2+2+∫01d(t+1t)(t+1t)2−2 =12tan−1t−1t2|01+122ln(t+1t−2t+1t+2)|01 =π22+122ln(2−12+2)=π22−12ln(2+1) a+b=122−12=−122