Evaluate the following integrals: x^2 (x^2+a^2)(x^2+b^2) dx - Vidyadip

Question

Evaluate the following integrals:
$\int\frac{\text{x}^2}{(\text{x}^2+\text{a}^2)(\text{x}^2+\text{b}^2)}\ \text{dx}$

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Answer

Let $\text{I}=\int\frac{1}{\sin\text{x}(3+2\cos\text{x})}\ \text{dx}$
$\int\frac{\sin\text{x dx}}{\sin^2\text{x}(3+2\cos\text{x})}$
$=\int\frac{\sin\text{x dx}}{(1-\cos^2\text{x})(3+2\cos\text{x})}$
Let $\cos\text{x}=\text{t}$
$\Rightarrow-\sin\text{x dx}=\text{dt}$
$\therefore\text{I}=\int\frac{\text{dt}}{(\text{t}^2-1)(3+2\text{t})}$
Now,
Let $\frac{1}{(\text{t}^2-1)(3+2\text{t})}=\frac{\text{A}}{\text{t}-1}+\frac{\text{B}}{\text{t}+1}+\frac{\text{C}}{3+2\text{t}}$
⇒ 1 = A(t + 1)(3 + 2t) + B (t - 1)(3 + 2t) + C (t² - t)
Put t = 1
⇒ 1 = 10A ⇒ $\text{A}=\frac{1}{10}$
Put t = -1
⇒ 1 = 2B ⇒ $\text{B}=-\frac{1}{2}$
Put $\text{t}=-\frac{3}{2}$
$\Rightarrow1=\frac{5}{4}\text{C}\Rightarrow\text{C}=\frac{4}{5}$
thus,
$\text{I}=\frac{1}{10}\int\frac{\text{dt}}{\text{t}-1}-\frac{1}{2}\int\frac{\text{dt}}{\text{t}+1}+\frac{5}{4}\int\frac{\text{dt}}{3+2\text{t}}$
$=\frac{1}{10}\log|\text{t}-1|-\frac{1}{2}\log|\text{t}+1|+\frac{2}{5}\log|3+2\text{t}|+\text{C}$
Hence,
$\text{I}=\frac{1}{10}\log|\cos\text{x}-1|-\frac{1}{2}\log|\cos\text{x}+1|+\frac{2}{5}\log|3+2\cos\text{x}|+\text{C}$

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