Integrals — Maths STD 12 Science — Question

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 x² + 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 I₁ and I₂ in eq. (iii),

$I = \sqrt {{x^2} + 2x + 3} + \log \left| {x + 1 + \sqrt {{x^2} + 2x + 3} } \right| + c$