First slide

Methods of integration

Question

If $f (x) = lim_{n \to \infty} [2 x + 4 x^{3} + \dots + 2 {nx}^{2 n - 1}] (0 < x < 1)$ , then $\int f (x) dx$ is equal to

Difficult

A

$- \sqrt{1 - x^{2}}$
B

$\frac{1}{\sqrt{1 - x^{2}}}$
C

$\frac{1}{x^{2} - 1}$
D

$\frac{1}{1 - x^{2}}$

Solution

Let, $\begin{array}{l} g_{n} (x) = 1 + x^{2} + x^{4} + \dots + x^{2 n} = \frac{x^{2 n + 2} - 1}{x^{2} - 1} \\ h_{n} (x) = g_{n}^{'} (x) = \frac{2 x ({nx}^{2 n + 2} - (n + 1) x^{2 n} + 1)}{{(x^{2} - 1)}^{2}} \end{array}$
Now $f (x) = lim_{n \to \infty} h_{n} (x) = \frac{2 x}{{(x^{2} - 1)}^{2}}$
As $0 < x < 1$
Thus, $\int f (x) dx = \int \frac{2 x}{{(x^{2} - 1)}^{2}} dx = - \frac{1}{x^{2} - 1} = \frac{1}{1 - x^{2}}$

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