Explained: "What is the integral of 11+x2?"

Shirley Thompson

Answered question

2021-12-21

What is the integral of

\frac{1}{1 + x^{2}}

abonirali59

Beginner2021-12-22Added 35 answers

\int \frac{1}{1 + x^{2}} d x = \tan^{- 1} x + C

Explanation:

\int \frac{d u}{1 + u^{2}} = \tan^{- 1} u + C \Rightarrow Where u is a function of x

Proof:

\int \frac{d u}{1 + u^{2}}

Integration by Trigonometric Substitution

u = \tan θ \Rightarrow d u = \sec^{2} θ d (θ)

\int \frac{d u}{1 + u^{2}} = \int \frac{\sec^{2} θ d (θ)}{1 + \tan^{2} θ}

\sec^{2} θ = 1 + \tan^{2} θ

\int \frac{\sec^{2} θ d (θ)}{1 + \tan^{2} θ} = \int \frac{(1 + \tan^{2} θ) d (θ)}{1 + \tan^{2} θ}

= \int d (θ) = θ

Reverse the Substitution

u = \tan θ \Rightarrow θ = \tan^{- 1} u

∴ \int \frac{d u}{1 + u^{2}} = \tan^{- 1} u + C

Simply by Substituting in this relation

\int \frac{d x}{1 + x^{2}} = \tan^{- 1} x + C

\arctan x + C

Jeffery Autrey

Beginner2021-12-23Added 35 answers

\arctan x + C .

Explanation:
It is one of the Standard Integral :

\int \frac{1}{1 + x^{2}} = \arctan x + C

Aliter :
Let,

I = \int \frac{1}{1 + x^{2}} d x

Subst.

x = \tan u . ∴ d x = \sec^{2} u d u, u = \arctan x

∴ I = \int \frac{1}{1 + \tan^{2} u} \sec^{2} u d u

= \int \frac{1}{\sec^{2} u} \sec^{2} u d u

= \int 1 d u

= u

\Rightarrow I = \arctan x + C

RizerMix

Expert2021-12-29Added 656 answers

I hope you can help me please

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