Q89 / 150 · MHT CET 2026 (19 April Shift 2) · Mathematics
If $f(x)$ is an antiderivative of $\dfrac{x + 1}{\sqrt{x - 1}}$ with respect to $x$, then value of $f(5) - f(2)$ is...
- A$4$
- B$\dfrac{14}{3}$
- C$7$
- Correct answerD$\dfrac{26}{3}$
Solution
$f(x) = \int \dfrac{x + 1}{\sqrt{x - 1}} dx$
$f(x) = \int \dfrac{x - 1 + 2}{\sqrt{x - 1}} dx$
$f(x) = \int \left( \sqrt{x - 1} + \dfrac{2}{\sqrt{x - 1}} \right) dx$
$f(x) = \dfrac{2}{3}(x - 1)^{3/2} + 4(x - 1)^{1/2} + C$
$f(5) = \dfrac{2}{3}(4)^{3/2} + 4(4)^{1/2} + C = \dfrac{16}{3} + 8 + C = \dfrac{40}{3} + C$
$f(2) = \dfrac{2}{3}(1)^{3/2} + 4(1)^{1/2} + C = \dfrac{2}{3} + 4 + C = \dfrac{14}{3} + C$
$f(5) - f(2) = \dfrac{40}{3} - \dfrac{14}{3} = \dfrac{26}{3}$
Answer: $\dfrac{26}{3}$
$f(x) = \int \dfrac{x - 1 + 2}{\sqrt{x - 1}} dx$
$f(x) = \int \left( \sqrt{x - 1} + \dfrac{2}{\sqrt{x - 1}} \right) dx$
$f(x) = \dfrac{2}{3}(x - 1)^{3/2} + 4(x - 1)^{1/2} + C$
$f(5) = \dfrac{2}{3}(4)^{3/2} + 4(4)^{1/2} + C = \dfrac{16}{3} + 8 + C = \dfrac{40}{3} + C$
$f(2) = \dfrac{2}{3}(1)^{3/2} + 4(1)^{1/2} + C = \dfrac{2}{3} + 4 + C = \dfrac{14}{3} + C$
$f(5) - f(2) = \dfrac{40}{3} - \dfrac{14}{3} = \dfrac{26}{3}$
Answer: $\dfrac{26}{3}$
Hence, the correct answer is option: 'D'.