∫ $\frac{5\mathit{x}+3}{\sqrt{{\mathit{x}}^2+4\mathit{x}+10}}$ dx
Now, 5x + 3 = A $\frac{\mathit{d}}{\mathit{d}\mathit{x}}\left({\mathit{x}}^2+4\mathit{x}+10\right)$ + 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
∫ $\frac{5\mathit{x}+3}{\sqrt{{\mathit{x}}^2+4\mathit{x}+10}}$ dx = ∫ $\frac{\left[\frac{5}{2}\frac{\mathit{d}}{\mathit{d}\mathit{x}}\left({\mathit{x}}^2+4\mathit{x}+10\right)-7\right]}{\sqrt{{\mathit{x}}^2+4\mathit{x}+10}}$ dx
= ∫ $\left[\frac{\frac{5}{2}\left(2\mathit{x}+4\right)-7}{\sqrt{{\mathit{x}}^2+4\mathit{x}+10}}\right]$ dx
= $\frac{5}{2}$ ∫ $\frac{2\mathit{x}+4}{\sqrt{{\mathit{x}}^2+4\mathit{x}+10}}$ dx - 7 ∫ $\frac{\mathit{d}\mathit{x}}{\sqrt{{\mathit{x}}^2+4\mathit{x}+10}}$
= $\frac{5}{2}{\mathit{I}}_1-7{\mathit{I}}_2$ ... (1)
${\mathit{I}}_1$ = ∫ $\frac{2\mathit{x}+4}{\sqrt{{\mathit{x}}^2+4\mathit{x}+10}}$ dx
Put ${\mathit{x}}^2$ + 4x + 10 = ${\mathit{z}}^2$
(2x + 4)dx = 2zdz
Thus, ${\mathit{I}}_1$ = ∫ $\frac{2\mathit{z}}{\mathit{z}}$ dz = 2z = 2 $\sqrt{{\mathit{x}}^2+4\mathit{x}+10}+{\mathit{C}}_1$
${\mathit{I}}_2$ = ∫ $\frac{\mathit{d}\mathit{x}}{\sqrt{{\mathit{x}}^2+4\mathit{x}+10}}$
= ∫ $\frac{\mathit{d}\mathit{x}}{\sqrt{{\mathit{x}}^2+4\mathit{x}+4+6}}$
= ∫ $\frac{\mathit{d}\mathit{x}}{\sqrt{{\left(\mathit{x}+2\right)}^2+{\left(\sqrt{6}\right)}^2}}$
= log |(x + 2) + $\sqrt{{\mathit{x}}^2+4\mathit{x}+10}$| + ${\mathit{C}}_2$
Substituting ${\mathit{I}}_1$ and ${\mathit{I}}_2$ in(1),we get
∴ ∫ $\frac{5\mathit{x}+3}{\sqrt{{\mathit{x}}^2+4\mathit{x}+10}}$ dx = $\frac{5}{2}\left(2\sqrt{{\mathit{x}}^2+4\mathit{x}+10}+{\mathit{C}}_1\right)$ - 7 [log |(x + 2) + $\sqrt{{\mathit{x}}^3+4\mathit{x}+10}$| + ${\mathit{C}}_2$]
= 5 $\sqrt{{\mathit{x}}^2+4\mathit{x}+10}$ - 7 [log |(x + 2) + $\sqrt{{\mathit{x}}^3+4\mathit{x}+10}$|] + $\frac{5}{2}{\mathit{C}}_1-7{\mathit{C}}_2$
= 5 $\sqrt{{\mathit{x}}^2+4\mathit{x}+10}$ - 7 [log |(x + 2) + $\sqrt{{\mathit{x}}^3+4\mathit{x}+10}$|] + C , where C = $\frac{5}{2}{\mathit{C}}_1-7{\mathit{C}}_2$
OR
I = ∫ $\frac{2\mathit{x}}{\sqrt{\left({\mathit{x}}^2+1\right)\left({\mathit{x}}^2+3\right)}}$
Let ${\mathit{x}}^2$ = z
∴ 2xdx = dz
∫ I = ∫ $\frac{\mathit{d}\mathit{z}}{\left(\mathit{z}+1\right)\left(\mathit{z}+3\right)}$
By partial fraction $\frac{1}{\left(\mathit{z}+1\right)\left(\mathit{z}+3\right)}$ = $\frac{\mathit{A}}{\mathit{z}+1}+\frac{\mathit{B}}{\mathit{z}+3}$
⇒ 1 = A (z + 3) + B (z + 1)
Putting z = - 3, we obtain :
1 = - 2B
B = - $\frac{1}{2}$
∴ A = $\frac{1}{2}$
∴ $\frac{1}{\left(\mathit{z}+1\right)\left(\mathit{z}+3\right)}$ = $\frac{\frac{1}{2}}{\mathit{z}+1}$ + $\frac{\left(-\frac{1}{2}\right)}{\mathit{z}+3}$
⇒ ∫ $\frac{\mathit{d}\mathit{z}}{\left(\mathit{z}+1\right)\left(\mathit{z}+3\right)}$ = $\frac{1}{2}$ ∫ $\frac{\mathit{d}\mathit{z}}{\mathit{z}+1}-\frac{1}{2}$ ∫ $\frac{\mathit{d}\mathit{z}}{\mathit{z}+3}$
= 1/2 log |z + 1| - $\frac{1}{2}$ log |z + 3| + C
∴ ∫ $\frac{2\mathit{x}\mathit{d}\mathit{x}}{\left({\mathit{x}}^2+1\right)\left({\mathit{x}}^2+3\right)}$ = $\frac{1}{2}$ log $\left|{\mathit{x}}^2+1\right|$ - $\frac{1}{2}$ log $\left|{\mathit{x}}^2+3\right|$ + C