∫ $\frac{5x+3}{\sqrt{x^2+4x+10}}$ dx Now, 5x + 3 = A $\frac{d}{dx}(x^2+4x+10)$ + B ⇒ 5x + 3 = A (2x + 4) + B ⇒ 5x + 3 = 2Ax + 4A + B ⇒ 2A = 5 and 4A + B = 3 ⇒ A = $\frac{5}{2}$ Thus, 4 $\left(\frac{5}{2}\right)$ + B = 3 ⇒ 10 + B = 3 ⇒ B = 3 - 10 = - 7 On substituting the values of A and B,we get ∫ $\frac{5x+3}{\sqrt{x^2+4x+10}}$ dx = ∫ $\frac{\frac{5}{2}\frac{d}{dx}(x^2+4x+10)-7}{\sqrt{x^2+4x+10}}$ dx = ∫ $\frac{\frac{5}{2}(2x+4)-7}{\sqrt{x^2+4x+10}}$ dx = $\frac{5}{2}$ ∫ $\frac{2x+4}{\sqrt{x^2+4x+10}}$ dx - 7 ∫ $\frac{dx}{\sqrt{x^2+4x+10}}$ = $\frac{5}{2}I_1-7I_2$ ... (1) $I_1$ = ∫ $\frac{2x+4}{\sqrt{x^2+4x+10}}$ dx Put $x^2$ + 4x + 10 = $z^2$ (2x + 4)dx = 2zdz Thus, $I_1$ = ∫ $\frac{2z}{z}$ dz = 2z = 2 $\sqrt{x^2+4x+10} + C_1$ $I_2$ = ∫ $\frac{dx}{\sqrt{x^2+4x+10}}$ = ∫ $\frac{dx}{\sqrt{x^2+4x+4+6}}$ = ∫ $\frac{dx}{\sqrt{(x+2)^2+(\sqrt{6})^2}}$ = log |(x + 2) + $\sqrt{x^2+4x+10}$| + $C_2$ Substituting $I_1$ and $I_2$ in(1),we get ∴ ∫ $\frac{5x+3}{\sqrt{x^2+4x+10}}$ dx = $\frac{5}{2}(2\sqrt{x^2+4x+10}+C_1)$ - 7 [log |(x + 2) + $\sqrt{x^3+4x+10}$| + $C_2$] = 5 $\sqrt{x^2+4x+10}$ - 7 [log |(x + 2) + $\sqrt{x^3+4x+10}$|] + $\frac{5}{2}C_1-7C_2$ = 5 $\sqrt{x^2+4x+10}$ - 7 [log |(x + 2) + $\sqrt{x^3+4x+10}$|] + C , where C = $\frac{5}{2}C_1- 7C_2$ OR I = ∫ $\frac{2x}{\sqrt{(x^2+1)(x^2+3)}}$ Let $x^2$ = z ∴ 2xdx = dz ∫ I = ∫ $\frac{dz}{(z+1)(z+3)}$ By partial fraction $\frac{1}{(z+1)(z+3)}$ = $\frac{A}{z+1} + \frac{B}{z+3}$ ⇒ 1 = A (z + 3) + B (z + 1) Putting z = - 3, we obtain : 1 = - 2B B = - $\frac{1}{2}$ ∴ A = $\frac{1}{2}$ ∴ $\frac{1}{(z+1)(z+3)}$ = $\frac{\frac{1}{2}}{z+1}$ + $\frac{-\frac{1}{2}}{z+3}$ ⇒ ∫ $\frac{dz}{(z+1)(z+3)}$ = $\frac{1}{2}$ ∫ $\frac{dz}{z+1}-\frac{1}{2}$ ∫ $\frac{dz}{z+3}$ = 1/2 log |z + 1| - $\frac{1}{2}$ log |z + 3| + C ∴ ∫ $\frac{2x\,dx}{(x^2+1)(x^2+3)}$ = $\frac{1}{2}$ log $\left|x^2+1\right|$ - $\frac{1}{2}$ log $\left|x^2+3\right|$ + C