Proving a standard integral? (why the integral of 1 / (1 + x^2) is equal to arctan x + c)

Skelly4444

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Apr 4, 2019
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In mathematics, if we're asked to prove a standard integral, where can we find step by step solutions, or are we just supposed to accept it as it stands?

I am trying to find out exactly why the integral of 1 / (1 + x^2) is equal to arctan x + c

Any advice would be greatly appreciated
 
It would be sufficient to show that:

ddx(arctan(x))=1x2+1\displaystyle \frac{d}{dx}(\arctan(x))=\frac{1}{x^2+1}

Let's begin with:

y=arctan(x)\displaystyle y=\arctan(x)

Or:

x=tan(y)\displaystyle x=\tan(y)

Differentiate with respect to x...what do you find?
 
Many thanks for replying.

I have had another look at this using the substitution x=tan u and I managed to integrate it and obtain the answer of arctan x in the end

It wasn't as difficult as I'd imagined it would be.
 
It would be sufficient to show that:

ddx(arctan(x))=1x2+1\displaystyle \frac{d}{dx}(\arctan(x))=\frac{1}{x^2+1}

Let's begin with:

y=arctan(x)\displaystyle y=\arctan(x)

Or:

x=tan(y)\displaystyle x=\tan(y)

Differentiate with respect to x...what do you find?

We get:

1=sec2(y)dydx\displaystyle 1=\sec^2(y)\frac{dy}{dx}

1sec2(y)=dydx\displaystyle \frac{1}{\sec^2(y)}=\frac{dy}{dx}

Apply a Pythagorean identity:

1tan2(y)+1=dydx\displaystyle \frac{1}{\tan^2(y)+1}=\frac{dy}{dx}

1x2+1=dydx\displaystyle \frac{1}{x^2+1}=\frac{dy}{dx}

Shown as desired.
 
You likely began with:

1x2+1dx\displaystyle \int\frac{1}{x^2+1}\,dx

Let:

x=tan(u)dx=sec2(u)du\displaystyle x=\tan(u)\rightarrow\,dx=\sec^2(u)\,du

sec2(u)tan2(u)+1du\displaystyle \int\frac{\sec^2(u)}{\tan^2(u)+1}\,du

Applying that same Pythagorean identity:

du=u+C=arctan(x)+C\displaystyle \int\,du=u+C=\arctan(x)+C
 
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