Question

If ${\displaystyle \int \frac{\text{dx}}{(x+2)(x^2+1)}=\text{a}\log |1+x^2|+\text{b}{\tan }^{-1}x+\frac{1}{5}\log |x+2|+\text{C}}$, then ______.

विकल्प

• a = ${\displaystyle \frac{-1}{10}}$, b = ${\displaystyle \frac{-2}{5}}$ 

• a = ${\displaystyle \frac{1}{10}}$, b = ${\displaystyle -\frac{2}{5}}$

• a = ${\displaystyle \frac{-1}{10}}$, b = ${\displaystyle \frac{2}{5}}$

• a = ${\displaystyle \frac{1}{10}}$, b = ${\displaystyle \frac{2}{5}}$

Answer 1

If ${\displaystyle \int \frac{\text{dx}}{(x+2)(x^2+1)}=\text{a}\log |1+x^2|+\text{b}{\tan }^{-1}x+\frac{1}{5}\log |x+2|+\text{C}}$, then a = ${\displaystyle \frac{-1}{10}}$, b = ${\displaystyle \frac{2}{5}}$.

Explanation:

Given that, ${\displaystyle \int \frac{\text{dx}}{(x+2)(x^2+1)}=\text{a}\log |1+x^2|+\text{b}{\tan }^{-1}x+\frac{1}{5}\log |x+2|+\text{C}}$

Now, I = ${\displaystyle \int \frac{\text{dx}}{(x+2)(x^2+1)}}$

${\displaystyle \frac{1}{(x+2)(x^2+1)}=\frac{\text{A}}{x+2}+\frac{\text{B}x+\text{C}}{x^2+1}}$

⇒ 1 = A(x² + 1) + (Bx + C)(x + 2)

⇒ 1 = (A + B)x² + (2B + C)x + A + 2C

Comapring coefficient, we get

A + B = 0

A + 2C = 1

2B + C = 0

Solving we get A = ${\displaystyle \frac{1}{5}}$

B = ${\displaystyle -\frac{1}{5}}$

And C = ${\displaystyle \frac{2}{5}}$

∴ ${\displaystyle \int \frac{\text{dx}}{(x+2)(x^2+1)}}$

= ${\displaystyle \frac{1}{5}\int \frac{1}{x+2}\text{d}x+\int \frac{-\frac{1}{5}+\frac{2}{5}}{x^2+1}\text{d}x}$

= ${\displaystyle \frac{1}{5}\int \frac{1}{x+2}\text{d}x-\frac{1}{10}\int \frac{2x}{1+x^2}\text{d}x+\frac{1}{5}\int \frac{2}{1+x^2}\text{d}x}$

= ${\displaystyle \frac{1}{5}\log |x+2|-\frac{1}{10}\log |1+x^2|+\frac{2}{5}{\tan }^{-1}x+\text{C}}$

= ${\displaystyle \text{a}\log |1+x^2|+\text{b}{\tan }^{-1}x+\frac{1}{5}\log |x+2|+\text{C}}$  ....(Given)

∴ a = ${\displaystyle \frac{-1}{10}}$, b = ${\displaystyle \frac{2}{5}}$.