Evaluate: 
OR
Evaluate: 




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I1 = ![]()
Let x2 + 4x + 10 = t
Differentiating w.r.t x
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![]()
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
![]()
Couldn't generate an explanation.
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