Q. 175.0( 3 Votes )

# Evaluate: <span l

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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