CUET Preparation Today
Which of the following is true ? |
$∫\frac{dx}{x^2+a^2}=\frac{1}{2a}log\left|\frac{x-a}{x+a}\right|+c,$ where c is an arbitrary constant $∫\frac{dx}{x^2-a^2}=\frac{1}{2a}log\left|\frac{x-a}{x+a}\right|+c,$ where c is an arbitrary constant $∫\frac{dx}{x^2-a^2}=\frac{1}{2a}log\left|\frac{x+a}{x-a}\right|+c,$ where c is an arbitrary constant $∫\frac{dx}{x^2+a^2}=\frac{1}{2a}log\left|\frac{x+a}{x-a}\right|+c,$ where c is an arbitrary constant |
$∫\frac{dx}{x^2-a^2}=\frac{1}{2a}log\left|\frac{x-a}{x+a}\right|+c,$ where c is an arbitrary constant |
The correct answer is Option (2) → $∫\frac{dx}{x^2-a^2}=\frac{1}{2a}log\left|\frac{x-a}{x+a}\right|+c,$ where c is an arbitrary constant $∫\frac{dx}{x^2+a^2}=\frac{1}{2a}\int\frac{(x+a)-(x-a)}{(x-a)(x+a)}dx$ $=\frac{1}{2a}\left[\int\frac{1}{x-a}dx-\int\frac{1}{x+a}dx\right]$ $=\frac{1}{2a}\left[\log|x-a|-\log|x+a|\right]$ $=\frac{1}{2a}\log\left|\frac{x-a}{x+a}\right|+c$ [c = constant] |
