TS EAMCET 4 May 2018 Shift 1 Solved Paper

Consider the function.
∫

x

(x2+1 )(x−1)

dx=A‌log‌|x2+1| +Btan−1x+C‌log‌|x−1|+d
Let

x

(x2+1)(x−1)

=

P

x−1

+

(Qx+R)

(x2+1)

Then,
x=P(x2+1)+(Qx+R)(x−1)
Compare the coefficient of like terms
P+Q=0
R−Q=1
P−R=0
Therefore,
P=

1

2

Q=−

1

2

R=

1

2

So,
I=∫

x

(x2+1)(x−1)

dx
=

1

2

‌∫

dx

x−1

+

1

2

‌∫

−x+1

x2+1

dx
=

1

2

‌log‌|x−1|−

1

4

|x2+1| +

1

2

tan−1(x)+d
Compare with the original equation.
A=−

1

4

B=

1

2

C=

1

2

Therefore,
A+B+C=−

1

4

+

1

2

+

1

2

=

3

4