Find the value of f(x) when: f(x) = _0^x dt/t^2 + 9 - Mathematics MCQ Mathematics Mock Test - 2 2026 | ExamTest.live

mediumMCQMathematics Mock Test - 22026Mathematics

1 attempts0% success rate1 mark

Find the value of f (x) when: f (x) = \int_{0}^{x} \frac{d t}{t ^{2} + 9}

Solution & Step-by-step Explanation

We use the standard integration formula: \int \frac{d x}{x ^{2} + a ^{2}} = \frac{1}{a} tan^{- 1} (\frac{x}{a}) + c In our case, a^{2} = 9 ⟹ a = 3 . f (x) = [\frac{1}{3} tan^{- 1} (\frac{t}{3})]_{0}^{x} f (x) = \frac{1}{3} tan^{- 1} (\frac{x}{3}) - \frac{1}{3} tan^{- 1} (0) Since tan^{- 1} (0) = 0 : f (x) = \frac{1}{3} tan^{- 1} (\frac{x}{3})

Try it yourself before checking the explanation above.

Find the value of f (x) when: f (x) = \int_{0}^{x} \frac{d t}{t ^{2} + 9}

\frac{1}{3} tan^{- 1} (\frac{x}{3})

\frac{1}{9} tan^{- 1} (\frac{x}{9})

tan^{- 1} (\frac{x}{3})

tan^{- 1} (\frac{x}{9})

Practice with timed mock tests and track your performance across Mathematics.