Question

if \(∫ \frac{(x^2+1)e^x}{(x+1)^2}dx = ƒ(x)e^x+C\) where \(C\) is a constant, then \(\frac{d^3ƒ}{dx^3}\) at \(x = 1\) is equal to :

• A. \(-\frac{3}{4}\)
• B. \(\frac{3}{4}\)
• C. \(-\frac{3}{2}\)
• D. \(\frac{3}{2}\)

Answer 1

The Correct Option is B

\(I = ∫ \frac{(x^2+1)e^x}{(x+1)^2}dx = ƒ(x)e^x+C\)

\(I = ∫ \frac{e^x(x^2-1+1+1)}{(x+1)^2}dx\)

= \(∫ e^x\bigg[\frac{x-1}{x+1}+\frac{2}{(x+1)^2}\bigg]dx\)

= \(e^x\bigg(\frac{x-1}{x+1}\bigg)+c\)

\(∴ f (x) = \frac{x-1}{x+1}\)

\(f(x) = \frac{1- 2}{x+1}\)

\(f' (x) = 2\bigg(\frac{1}{x+1}\bigg)^2\)

\(f''(x) = -4\bigg(\frac{1}{x+1}\bigg)^3\)

\(f'''(x) = \frac{12}{(x+1)^4}\)

for \(x = 1\)

\(f'''(1) = \frac{12}{24}\)

= \(\frac{12}{16}\)

= \(\frac{3}{4}\)

Hence, the correct option is (B): \(\frac{3}{4}\)