Reciprocal quadratic integral

Rewrite the integral as $\int _{ -1 }^{ \sqrt { 2 } }{ \frac { 5 }{ 25{ x }^{ 2 }+70x+49+1 } } dx\\ =\int _{ -1 }^{ \sqrt { 2 } }{ \frac { 5 }{ { (5x+7) }^{ 2 }+1 } dx }$ .

Now let $u=5x+7$

$\rightarrow du=5dx$

When $x=-1, u=2$ and when $x=\sqrt {2}, u=5\sqrt {2} +7$ .

Hence the integral is $\int _{ 2 }^{ 5\sqrt { 2 } +7 }{ \frac { 1 }{ 1+{ u }^{ 2 } } } du\\ ={ \left[ \tan ^{ -1 }{ u } \right] }_{ 2 }^{ 5\sqrt { 2 } +7 }\\ =\tan ^{ -1 }{ (5\sqrt { 2 } +7)-\tan ^{ -1 }{ (2) } }$ .

Now let $A=\tan ^{ -1 }{ (5\sqrt { 2 } +7) }$ and $B=\tan ^{ -1 }{ (2) }$ . Taking tan of both sides:

$\tan { (A-B)=\frac { \tan { A } -\tan { B } }{ 1+\tan { A } \tan { B } } } \\ =\frac { 5\sqrt { 2 } +7-2 }{ 1+(5\sqrt { 2 } +7)2 } \\ =\frac { 5\sqrt { 2 } +5 }{ 10\sqrt { 2 } +15 } \\ =\frac { \sqrt { 2 } +1 }{ 2\sqrt { 2 } +3 } \\ =\sqrt { 2 } -1$ .

Hence $A-B=\tan ^{ -1 }{(\sqrt {2} -1)}$ . To find this in the required form, consider $\tan {2(A-B)}$ :

$\tan { 2(A-B) } =\frac { 2\tan { (A-B) } }{ 1-\tan ^{ 2 }{ (A-B) } } \\ =\frac { 2\sqrt { 2 } -2 }{ 2\sqrt { 2 } -2 } \\ =1.$

$2(A-B)=\tan ^{ -1 }{ 1 } \\ =\frac { \pi }{ 4 } \\ \therefore \quad A-B=\frac { \pi }{ 8 } \\ a=1,\quad b=8,\quad a+b=\boxed { 9 }$