The integral of tan^(−1)(x) is - A. x tan^(−1)(x)−x ln|1+x^2|+c B. x^2 tan^(−1)(x)−1/2 ln|1+x^2|+c C. x tan^(−1)(x)−1/2 ln|1+x^2|+c D. x tan^(−1)(x)−x/2 ln|1+x^2|+c

sublimnoj7u7

sublimnoj7u7

Answered question

2023-01-01

The integral of tan 1 ( x ) is -
A. x tan 1 ( x ) x ln | 1 + x 2 | + c
B. x 2 tan 1 ( x ) 1 2 ln | 1 + x 2 | + c
C. x tan 1 ( x ) 1 2 ln | 1 + x 2 | + c
D. x tan 1 ( x ) x 2 ln | 1 + x 2 | + c

Answer & Explanation

Rihanna Gamble

Rihanna Gamble

Beginner2023-01-02Added 12 answers

tan 1 ( x ) d x = ( tan 1 ( x ) × 1 ) d x
= tan 1 ( x ) 1 d x [ d d x ( tan 1 ( x ) 1 d x ] d x
= tan 1 ( x ) x 1 1 + x 2 x d x
= x tan 1 ( x ) 1 2 2 x 1 + x 2 d x
During the second term, if a replacement, t = 1 + x 2 , d t = 2 x d x, which is the numerator. So we get
tan 1 ( x ) d x = x tan 1 ( x ) 1 2 d t t
= x tan 1 ( x ) 1 2 ln | t |
Substituting back t = 1 + x 2
tan 1 ( x ) d x = x tan 1 ( x ) 1 2 ln | 1 + x 2 | + c

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?