Q. 175.0( 3 Votes )

Evaluate: <span l

Answer :






I1 =


Let x2 + 4x + 10 = t


Differentiating w.r.t x




Therefore, I1 =


Putting the values of
& x2 + 4x + 10 = t


I1 =


=


=


=


=


=


=


=


=


=


As


Therefore,


I2 =


= =


Putting the values of I1 & I2,


I = I1 + I2



Hence


OR


Putting x2 = t


Differentiating


2x dx = dt



Let


1 = A(3 + t) + B (1 + t)


On Comparing,


1 = 3A + B


0 = A + B


On solving the above mentioned equations,


A = 0.5, B = -0.5




= 0.5 log (1 + t) – 0.5 log (3 + t) + C


= 0.5 log (1 + x2) – 0.5 log (3 + x2) + C



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