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
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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