0 x 1 + t2 dt Solution put t= tanu dt = sec2 u du secu . sec2u du sec3u du with limits 0 to x = 1/2 tan u .sec u + 1/2 ln (sec u + tan u) = 1/2 [ tan (tan-1 t) . sec (tan-1 t ) ] + 1/2 [ ln ( sec (tan-1 t) + tan (tan-1 t)] with limits 0 to x = 1/2 [ t . sec (sec-1(t2+1) ) ] + 1/2 [ ln (sec(sec-1 (t2 +1) ) + t ] with limits 0 to x = 1/2 [ t. (t2 +1) ] + 1/2 [ln( (t2 +1) ) + t) ] with limits 0 to x By applying limits we get 1/2 [( x.(x2+1)- 0 ] + 1/2 [ ln (x + (x2+1) - ln (1) ] = 1/2 [( x.(x2+1) ] + 1/2 [ ln (x + (x2+1)].