TS EAMCET 5 May 2018 Shift 2 Solved Paper

Consider the integral.
∫x(1+x)‌log(1+x2)‌d‌x= (F(x)‌log(1+x2)−

2

3

tan−1x−

2x3

9

−

x2

2

+

2x

3

+c)
Let,
I=∫x(1+x)‌log(1+x2)‌d‌x
=∫(x+x2)‌log(1+x2)‌d‌x
Solve the above integration by parts,
I=log(1+x2)(

x2

2

+

x3

3

) −∫

1

1+x2

.2x(

x2

2

+

x3

3

)dx
=log(1+x2)(

x2

2

+

x3

3

)−∫

x3

1+x2

dx −

2

3

‌∫(

x4

1+x2

)dx
Let I1=∫

x.x2

1+x2

dx
Let
x2=t
2xdx=dt
I1=

1

2

‌∫

t

1+t

dt
=

1

2

t−

1

2

‌log|1+t|+c1
=

1

2

x2−

1

2

‌log(1+x2)+c1
Let I2=∫

x4

1+x2

dx
∫

x4

1+x2

dx=∫

x4−1+1

1+x2

dx
=∫(

x2−1

1+x2

+

1

1+x2

)dx
=

x3

3

−x+tan−1x+c2
Therefore,
I=[log(1+x2)(

x2

2

+

x3

3

) −

1

2

x2+

1

2

‌log(1+x2)−

2

9

x3+

2

3

x−

2

3

tan−1x+c]
I=[log(1+x2)(

x2

2

+

x3

3

+

1

2

) −

2

3

tan−1x−

2

9

x3−

x2

2

+

2

3

x+C]
Comparing with original equation, we get
F(x)=

x2

2

+

x3

3

+

1

2