Chapter 5: Q 91. (page 431)

Prove the integration formula.
$鈭� \tan^{- 1} x d x = x \tan^{- 1} x - \frac{1}{2} \ln (x^{2} + 1) + C$
(a) by applying integration by parts to $鈭� \tan^{- 1} x d x$ .
(a) by differentiating $x \tan^{- 1} x - \frac{1}{2} \ln (x^{2} + 1)$ .

Short Answer

Expert verified

Part (a). The solution is $x \tan^{- 1} x - \frac{1}{2} \ln (x^{2} + 1) + C$ .

Part (b). The solution is $\tan^{- 1} x$ .

Step by step solution

Part (a) Step 1. Given information.

The given integral is:

$鈭� \tan^{- 1} x d x = x \tan^{- 1} x - \frac{1}{2} \ln (x^{2} + 1) + C$

Part (a) Step 2. Prove the formula by applying integration by parts.

Take $u = \tan^{- 1} x$ and $v = 1$ .

$\begin{array}{rcl} 鈭� \tan^{- 1} x 路 1 d x & = & \tan^{- 1} x 鈭� d x - 鈭� [\frac{d}{d x} (\tan^{- 1} x) 鈭� d x] d x \\ = & x 路 \tan^{- 1} x - 鈭� \frac{x}{1 + x^{2}} d x \\ = & x 路 \tan^{- 1} x - \frac{1}{2} 鈭� \frac{1}{t} d t 鈰� 鈰� 鈰� 鈰� 鈰� 鈰� \{p u t 1 + x^{2} = d t, 2 x d x = d t, x d x = \frac{1}{2} d t\} \\ = & x 路 \tan^{- 1} x - \frac{1}{2} \ln (t) + C \\ = & x 路 \tan^{- 1} x - \frac{1}{2} \ln (x^{2} + 1) + C \end{array}$

Hence, the formula is proved.

Part (b) Step 1. Given information.

The given formula is $鈭� \tan^{- 1} x d x = x 路 \tan^{- 1} x - \frac{1}{2} \ln (x^{2} + 1) + C$ .

Part (b) Step 2. Prove the formula by differentiating x·tan-1x-12lnx2+1.

$\begin{array}{rcl} \frac{d}{d x} \{x 路 \tan^{- 1} x - \frac{1}{2} \ln (x^{2} + 1)\} & = & x \frac{d}{d x} \{\tan^{- 1} x\} + \tan^{- 1} x \frac{d}{d x} \{x\} - \frac{1}{2} \frac{d}{d x} \{\ln (x^{2} + 1)\} \\ = & \frac{x}{1 + x^{2}} + \tan^{- 1} x - \frac{1}{2} 路 \frac{1}{1 + x^{2}} 路 2 x 鈰� 鈰� 鈰� 鈰� \{a p p l y c h a i n r u l e o f d i f f e r e n t i a t i o n\} \\ = & \frac{x}{1 + x^{2}} + \tan^{- 1} x - \frac{x}{1 + x^{2}} \\ = & \tan^{- 1} x \end{array}$

Hence, the formula is proved.

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91影视

Prove the integration formula.
$鈭� \tan^{- 1} x d x = x \tan^{- 1} x - \frac{1}{2} \ln (x^{2} + 1) + C$
(a) by applying integration by parts to $鈭� \tan^{- 1} x d x$ .
(a) by differentiating $x \tan^{- 1} x - \frac{1}{2} \ln (x^{2} + 1)$ .

Short Answer

Step by step solution

Part (a) Step 1. Given information.

Part (a) Step 2. Prove the formula by applying integration by parts.

Part (b) Step 1. Given information.

Part (b) Step 2. Prove the formula by differentiating x·tan-1x-12lnx2+1.

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91影视

Prove the integration formula.鈭�tan-1xdx=xtan-1x-12lnx2+1+C(a) by applying integration by parts to 鈭�tan-1xdx.(a) by differentiatingxtan-1x-12lnx2+1.

Short Answer

Step by step solution

Part (a) Step 1. Given information.

Part (a) Step 2. Prove the formula by applying integration by parts.

Part (b) Step 1. Given information.

Part (b) Step 2. Prove the formula by differentiating x·tan-1x-12lnx2+1.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Mechanics Maths

Geometry

Pure Maths

Statistics

Theoretical and Mathematical Physics

Discrete Mathematics

Study anywhere. Anytime. Across all devices.

Prove the integration formula.
$鈭� \tan^{- 1} x d x = x \tan^{- 1} x - \frac{1}{2} \ln (x^{2} + 1) + C$
(a) by applying integration by parts to $鈭� \tan^{- 1} x d x$ .
(a) by differentiating $x \tan^{- 1} x - \frac{1}{2} \ln (x^{2} + 1)$ .