(1 – x + x 2 ) dx = x tan –1 x – 
log (1 + x 2 ) + A, Then A is equal to -
log {1 + (1 – x) 2 } + c
log {1 + (1 – x) 2 } + c
log {1 + (1 – x) 2 } + c
Text Solution
Verified by ExpertsC
We have, 
(1 – x + x 2 ) dx
= 
{1 – x (1 – x)} dx
= 
dx
= 
dx
= 
x + tan –1 (1 – x)} dx
= 
x dx + 
(1 – x)dx
= I 1 + I 2 … (1)
where I 1 = 
x dx and I 2 = 
(1 – x) dx.
Now, I 1 = 
x dx= 
x 1 dx
I II
= x tan –1 x – 
dx
= x tan –1 x – 
d (1 + x 2 )
= x tan –1 x – 
log (1 + x 2 )… (2) and
I 2 = 
(1 – x) dx= – 
(1 – x) d (1 – x)
= – 
[using (2)]
Substituting the values of I 1 and I 2 in (1), we get
(1 – x + x 2 ) dx = x tan –1 x – 
log (1 + x 2 )
– (1 – x) tan –1 (1 – x) + 
log {1 + (1 – x) 2 } + c
Hence is the correct answer.
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