Let's do some calculus! (5)

Calculus Level 4

$\begin{aligned} f(x) & =\dfrac{x}{{\left(1+x^{n}\right)}^{{1}/{n}}} & \forall ~n \ge 2 \\ g(x) & =\underbrace{f \circ f \circ f \circ \cdots f}_{f \text{ occurs } n \text{ times}}\left(x\right) \end{aligned}$

For functions $f$ and $g$ as defined above, which of the following is an antiderivative of $\displaystyle \int x^{n-2} g(x) \, dx$ ?

Clarification : $C$ denotes the arbitrary constant of integration .

For more problems on calculus, click here .

\dfrac {\left(1+nx^n\right)^{1-\frac 1n}}{n(n-1)} + C

\dfrac{\left(1+nx^n\right)^{1+\frac{1}{n}}}{n(n+1)} + C

\dfrac{\left(1+nx^n\right)^{1+\frac{1}{n}}}{n+1} + C

\dfrac{\left(1+nx^n\right)^{1-\frac{1}{n}}}{n-1} + C

1 solution

Chew-Seong Cheong
Sep 19, 2016

$\begin{aligned} f(x) & = \frac x{(1+x^n)^\frac 1n} \\ \implies (f(x))^n & = \frac {x^n}{1+x^n} \\ (f \circ f (x))^n & = \frac {(f(x))^n}{1+(f(x))^n} = \frac {\frac {x^n}{1+x^n}}{1+\frac {x^n}{1+x^n}} = \frac {x^n}{1+2x^n} \\ (f \circ f \circ f (x))^n & = \frac {(f \circ f(x))^n}{1+(f \circ f(x))^n} = \frac {\frac {x^n}{1+2x^n}}{1+\frac {x^n}{1+\color{#D61F06}{2}x^n}} = \frac {x^n}{1+\color{#D61F06}{3}x^n} \\ \cdots \quad & = \quad \cdots \\ \implies (g(x))^n & = \frac {x^n}{1+\color{#D61F06}{n}x^n} \\ g(x) & = \frac x{(1+nx^n)^\frac 1n} \\ \int x^{n-2} g(x) \ dx & = \int \frac {x^{n-1}}{(1+nx^n)^\frac 1n} dx \\ & = \boxed{\dfrac {(1+nx^n)^{1-\frac 1n}}{n(n-1)}+C} \end{aligned}$

Tapas, I have edited your problem and the answer choices. Go through the LaTex codes. They can be simpler.

Chew-Seong Cheong - 4 years, 9 months ago

Where can I find them?

Tapas Mazumdar - 4 years, 8 months ago

If you are using a PC, place your mouse cursor on top of the formulas. Or at top right hand pull-down menu select Toggle LaTex.

Chew-Seong Cheong - 4 years, 8 months ago

@Chew-Seong Cheong – Yeah, I got it. Thanks!

Tapas Mazumdar - 4 years, 8 months ago

Let's do some calculus! (5)

For more problems on calculus, click here .

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