Mathematics

Find $$\int \dfrac{x+3}{\sqrt{5-4x+x^{2}}}dx$$.


SOLUTION
$$I = \int {\dfrac {{x + 3}}{{\sqrt {{x^2} - 4x + 5} }}\,dx} $$
$$Putting\,\,x + 3 = A\dfrac {d}{{dx}}\left( {{x^2} - 4x + 5} \right) + B$$
$$ = x + 3 = A\left( {2x - 4} \right) + B$$
$$ = x + 3 = 2Ax - 4A + B$$
$$On\,comparing\,both\,sides\,we\,get$$
$$2A = 1\,\, - 4A + B = 3$$
$$A = \dfrac {1}{2}\, \Rightarrow  - 4 \times \dfrac {1}{2} + B = 3$$
$$ \Rightarrow B - 2 = 3$$
$$ \Rightarrow B = 5$$
$$I = \int {\dfrac {{\dfrac {1}{2}\left( {2x - 4} \right) + 5}}{{\sqrt {{x^2} - 4x + 5} }}\,dx} \\$$
$$=\dfrac {1}{2}\int {\dfrac {{\left( {2x - 4} \right)dx}}{{\sqrt {{x^2} - 4x + 5} }}}  + 5\int {\dfrac {{dx}}{{\sqrt {{x^2} - 4x + 5} }}}$$
Put $$x^2-4x+5=t$$ 
thus,
$$(2x-4)dx=dt$$
$$= \dfrac {1}{2}\int {\dfrac {{dt}}{{\sqrt t }} + 5} \,\int {\dfrac {{dx}}{{\sqrt {{x^2} - 4x + 4 - 4 + 5} }}}$$
$$ = \dfrac {1}{2} \times 2\,{t^{\dfrac {1}{2}}} + 5\,\,\int {\dfrac {{dx}}{{{{\left( {x - 2} \right)}^2} + 1}}}$$
 $$= \sqrt t \, + \,5\,\log \,\left| {x - 2 + \sqrt {{{\left( {x - 2} \right)}^2} + 1} } \right| + c$$
 $$= \sqrt {{x^2} - 4x + 5}  + \,5\log \left| {x - 2 + \sqrt {{x^2} - 4x + 5} } \right| + c$$
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Subjective Medium Published on 17th 09, 2020
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