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Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIntegrals5 Marks
Question
Integrate the function $ \frac{{x + 2}}{{\sqrt {{x^2} + 2x + 3} }}$
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Answer

Let $I = \int {\frac{{x + 2}}{{\sqrt {{x^2} + 2x + 3} }}dx} $Let Linear = A $\frac{d}{{dx}}$ (Quadratic) + B
$\Rightarrow x + 2 = A\frac{d}{{dx}}\left( {{x^2} + 2x + 3} \right) + B$
$ \Rightarrow $ x + 2 = A(2x + 2) + B …(ii)
$ \Rightarrow $ x + 2 = 2Ax + 2A + B
Comparing coefficients of x, 2A = 1 $ \Rightarrow A = \frac{1}{2}$
Comparing constants, 2A + B = 2
On solving, we get $A = \frac{1}{2},$ B = 1
Putting the values of A and B in eq. (ii),
$x + 2 = \frac{1}{2}\left( {2x + 2} \right) + 1$
Putting this value of x+2 in eq. (i),
$I = \int {\frac{{\frac{1}{2}\left( {2x + 2} \right) + 1}}{{\sqrt {{x^2} + 2x + 3} }}dx} $
$I = \frac{1}{2}\int {\frac{{2x + 2}}{{\sqrt {{x^2} + 2x + 3} }}dx + \int {\frac{1}{{\sqrt {{x^2} + 2x + 3} }}dx} } $
$\Rightarrow I = \frac{1}{2}{I_1} + {I_2}$ …(iii)
Now ${I_1} = \int {\frac{{2x + 2}}{{\sqrt {{x^2} + 2x + 3} }}} dx$
Putting $x^2 + 2x + 3 = t$
$ \Rightarrow 2x + 2 = \frac{{dt}}{{dx}}$
$\Rightarrow $ (2x + 2)dx = dt
$\therefore {I_1} = \int {\frac{{dt}}{{\sqrt t }} = \int {{t^{\frac{{ - 1}}{2}}}dt} } $
$= \frac{{{t^{\frac{1}{2}}}}}{{\frac{1}{2}}} = 2\sqrt t $
$ = 2\sqrt {{x^2} + 2x + 3} $ …(iv)
Again ${I_2} = \int {\frac{1}{\sqrt{{x^2} + 2x + 3}}dx} $
$= \int {\frac{1}{{\sqrt {{x^2} + 2x + 1 + 2} }}} $
$= \int {\frac{1}{{\sqrt {{{\left( {x + 1} \right)}^2} + {{\left( {\sqrt 2 } \right)}^2}} }}dx} $
$= \log \left| {x + 1 + \sqrt {{{\left( {x + 1} \right)}^2} + {{\left( {\sqrt 2 } \right)}^2}} } \right|$
$= \log \left| {x + 1 + \sqrt {{x^2} + 2x + 3} } \right|$…(v)
Putting values of $I_1$ and $I_2$ in eq. (iii),
$I = \sqrt {{x^2} + 2x + 3} + \log \left| {x + 1 + \sqrt {{x^2} + 2x + 3} } \right| + c$

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