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Indefinite Integrals — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integrals5 Marks
Question
Evaluate the following intregals:
$\int\frac{4\text{x}^2+3}{(\text{x}^2+2)(\text{x}^2+3)(\text{x}^2+4)}\ \text{dx}$
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Answer

We have,
$\int\frac{(4\text{x}^2+3)\text{dx}}{(\text{x}^2+2)(\text{x}^2+3)(\text{x}^2+4)}\ $
Putting $x^2 = t$
Then,
$\frac{4\text{x}^2+3}{(\text{x}^2+2)(\text{x}^2+3)(\text{x}^2+4)}=\frac{4\text{t}^2+3}{(\text{t}+2)(\text{t}+3)(\text{t}+4)}$
Let $\frac{4\text{x}^2+3}{(\text{x}^2+2)(\text{x}^2+3)(\text{x}^2+4)}=\frac{\text{A}}{\text{t}+2}+\frac{\text{B}}{\text{t}+3}+\frac{\text{C}}{\text{t}+4}$
$\Rightarrow\frac{4\text{x}^2+3}{(\text{x}^2+2)(\text{x}^2+3)(\text{x}^2+4)}=\frac{\text{A}(\text{t}+3)(\text{t}+4)+\text{C}(\text{t}+2)(\text{t}+3)}{(\text{t}+2)(\text{t}+3)(\text{t}+4)}$
$\Rightarrow 4t^2 + 3 = A(t + 3)(t+ 4 ) + B(t + 2)(t + 4) + C(t + 2)(t + 3)$
putting $t + 3 = 0$
$⇒ t = -3$
$\therefore$ $4 \times (-3)^2 + 3 = B(-3 + 2)(-3 + 4)$
$\Rightarrow 39 = B(-1)$
$\Rightarrow B = -39$
Putting$ t + 2 = 0$
$⇒ t = -2$
$\therefore$ $4(-2)^2 + 3 = A(-2 + 3)(-2 + 4)$
$\therefore$ $19 = A \times 1 \times 2$
$\Rightarrow\text{A}=\frac{19}{2}$
Let $t + 4 = 0$
$\Rightarrow t = -4$
$\therefore$ $4 \times (-4)^2 + 3 = C(-4 + 2)(-4 + 3)$
$\Rightarrow 67 = C(-2)(-1)$
$\Rightarrow\text{C}=\frac{67}{2}$
$\therefore\frac{4\text{t}^2+3}{(\text{t}+2)(\text{t}+3)(\text{t}+4)}=\frac{19}{2(\text{t}+2)}-\frac{39}{\text{x}^2+3}-\frac{39}{(2\text{t}+4)}$
$\Rightarrow\frac{4\text{t}^2+3}{(\text{x}^2+2)(\text{x}^2+3)(\text{x}^2+4)}=\frac{19}{2(\text{x}^2+2)}-\frac{39}{\text{x}^2+3}+\frac{67}{2(\text{x}^2+4)}$
$\therefore\text{I}=\frac{19}{2}\int\frac{\text{dx}}{\text{x}^2+(\sqrt{2})^2}-39\int\frac{\text{dx}}{\text{x}^2+(\sqrt{3})^2}-\frac{67}{2}\int\frac{\text{dx}}{\text{x}^2+2^2}$
$=\frac{19}{2}\times\frac{1}{\sqrt{2}}\tan^{-1}\Big(\frac{\text{x}}{\sqrt{2}}\Big)-\frac{39}{\sqrt{3}}\tan^{-1}\Big(\frac{\text{x}}{\sqrt{3}}\Big)-67\times\frac{1}2{}\tan^{-1}\Big(\frac{\text{x}}{2}\Big)+\text{C}$
$=\frac{19}{2\sqrt{2}}\tan^{-1}\Big(\frac{\text{x}}{\sqrt{2}}\Big)-\frac{39}{\sqrt{3}}\tan^{-1}\Big(\frac{\text{x}}{\sqrt{3}}\Big)-\frac{67}{4}\tan^{-1}\Big(\frac{\text{x}}{2}\Big)+\text{C}$

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