First slide

Introduction to integration

Question

If $\int \frac{x \tan^{- 1} x}{\sqrt{1 + x^{2}}} d x =$
$\sqrt{1 + x^{2}} f (x) + K \log (x + \sqrt{x^{2} + 1}) + C$ then

Moderate

A

$f (x) = \tan^{- 1} x, K = - 1$
B

$f (x) = \tan^{- 1} x, K = 1$
C

$f (x) = 2 \tan^{- 1} x, K = - 1$
D

$f (x) = 2 \tan^{- 1} x, K = 1$

Solution

$\int \frac{x}{\sqrt{1 + x^{2}}} d x$
$= \frac{1}{2} \int \frac{d t}{\sqrt{t}} = \frac{1}{2} \cdot \frac{\sqrt{t}}{1 / 2} = \sqrt{t} = \sqrt{1 + x^{2}}$
$[w h e r e 1 + x 2 = t]$
we integrate by the parts, taking $\tan^{- 1} x$ as the first function and
$x / \sqrt{1 + x^{2}}$ as the second.
$\Rightarrow \begin{array}{l} \int \frac{x \tan^{- 1} x}{\sqrt{1 + x^{2}}} d x \\ = (\tan^{- 1} x) \sqrt{1 + x^{2}} - \int \frac{1}{1 + x^{2}} \cdot \sqrt{1 + x^{2}} d x \\ = \sqrt{1 + x^{2}} \tan^{- 1} x - \int \frac{d x}{\sqrt{1 + x^{2}}} \\ = \sqrt{1 + x^{2}} \tan^{- 1} x - \log (x + \sqrt{x^{2} + 1}) + C \end{array}$
Thus $f (x) = \tan^{- 1} x, K = - 1 .$

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