Integration Month

The question is wrong!!! I'm lucky 12 worked! :-D Following is the solution of the given integral:

$\int \frac { \arctan { (x } ) }{ { x }^{ 4 } } dx\\$

Using integration by parts:

$\arctan { (x) } \int { { x }^{ -4 } } dx\quad -\quad \int { \frac { 1 }{ 1+{ x }^{ 2 } } (\int { { x }^{ -4 }dx)\quad dx } } \\ \Rightarrow \quad \frac { -\arctan { (x) } }{ 3{ x }^{ 3 } } +\frac { 1 }{ 3 } \int { \frac { { x }^{ -3 } }{ 1+{ x }^{ 2 } } } dx\\ \Rightarrow \quad \frac { -\arctan { (x) } }{ 3{ x }^{ 3 } } +\frac { 1 }{ 3 } \int { \frac { { \quad }{ x }^{ -2 }{ x }^{ -3 } }{ 1+{ x }^{ -2 } } } dx$

Now, substituting ${ x }^{ -2 }=t\\ \Rightarrow \quad -2{ x }^{ -3 }dx\quad =\quad dt$

$\Rightarrow \quad \frac { -\arctan { (x) } }{ 3{ x }^{ 3 } } -\frac { 1 }{ 6 } \int { \frac { t-1 }{ t } dt } \\ \Rightarrow \quad \frac { -\arctan { (x) } }{ 3{ x }^{ 3 } } -\frac { 1 }{ 6 } \int { 1-\frac { 1 }{ t } dt } \\ \Rightarrow \quad \frac { -\arctan { (x) } }{ 3{ x }^{ 3 } } -\frac { 1 }{ 6 } t+\frac { 1 }{ 6 } \ln { t } +K\\ \Rightarrow \quad \frac { -\arctan { (x) } }{ 3{ x }^{ 3 } } -\frac { 1 }{ 6 } ({ x }^{ -2 }+1)+\frac { 1 }{ 6 } \ln { \frac { { x }^{ 2 }+1 }{ { x }^{ 2 } } } +K\\ \Rightarrow \quad \frac { -\arctan { (x) } }{ 3{ x }^{ 3 } } +\frac { 1 }{ 6 } \ln { \frac { { x }^{ 2 }+1 }{ { x }^{ 2 } } } -\frac { 1 }{ 6{ x }^{ 2 } } +C$

Now, by comparing, we can see, $a=3, b=1, d=2$ but $c$ can be both $6$ and $-6$ !

yeah i too got a=3 b=1 d=2 and c=6 and -6, which is not possible...... So this ques is wrongly stated!!!!!!!!!!