Integrals — Class 12 Maths Solution

: Let $I = \int {\cfrac{{5x + 3}}{{\sqrt {{x^2} + 4x + 10} }}dx}$

We write $5x + 3 = A\cfrac{d}{{dx}}\left( {{x^2} + 4x + 10} \right) + B$

$\Rightarrow$ $5x + 3 = 2xA + 4A + B$

….(i)
Comparing the coefficients of $x$ in (i),

we get
$5 = 2A$ $\Rightarrow$ $A = 5/2$
Comparing the constant terms in (i),

we get
$3 = 4A + B$ $\Rightarrow$ $B = - 7$

$\therefore$ $I = \int {\cfrac{{\cfrac{5}{2}\left( {2x + 4} \right) + \left( { - 7} \right)}}{{\sqrt {{x^2} + 4x + 10} }}} dx$

$= \cfrac{5}{2}\int {\cfrac{{2x + 4}}{{\sqrt {{x^2} + 4x + 10} }}dx} - 7\int {\cfrac{{dx}}{{\sqrt {{x^2} + 4x + 10} }}} = \cfrac{5}{2}{I_1} - 7{I_2}$

(say)
$\Rightarrow$ $I = \cfrac{5}{2}{I_1} - 7{I_2}$

….(ii)
Now, ${I_1} = \int {\cfrac{{2x + 4}}{{\sqrt {{x^2} + 4x + 10} }}dx}$

Put ${x^2} + 4x + 10 = t$ $\Rightarrow$ $\left( {2x + 4} \right)dx = dt$

$\therefore$ ${I_1} = \int {\cfrac{{dt}}{{\sqrt t }}} = \int {{t^{ - 1/2}}} dt = 2\sqrt t = 2\sqrt {{x^2} + 4x + 10} + {C_1}$

….(iii)
And ${I_2} = \int {\cfrac{{dx}}{{\sqrt {{x^2} + 4x + 10} }} = \int {\cfrac{{dx}}{{\sqrt {{{\left( {x + 2} \right)}^2} + {{\left( {\sqrt 6 } \right)}^2}} }}} }$

$= \log \left| {x + 2 + \sqrt {{{\left( {x + 2} \right)}^2} + {{\left( {\sqrt 6 } \right)}^2}} } \right|$

$= \log \left| {x + 2 + \sqrt {{x^2} + 4x + 10} } \right| + {C_2}$

….(iv)

Hence, from (ii), (iii) and (iv)

we get
$I = 5\sqrt {{x^2} + 4x + 10} - 7\log \left| {x + 2 + \sqrt {{x^2} + 4x + 10} } \right| + C$

$\left[ {C = \cfrac{5}{2}{C_1} - 7{C_2}} \right]$

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