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 Rate this question :

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