What is the method of integrating functions like 1/(1 + x^2)?
here's some more u should know by heart
$\displaystyle \displaystyle \int x^n \; dx = \frac {x^{n+1}}{n+1} + C $
$\displaystyle \displaystyle \int \sqrt[n]x \; dx = \frac {n}{n+1}x \sqrt[n]x + C $
$\displaystyle \displaystyle \int \frac {dx}{1+x^2 } =\arctan {x} + C $
$\displaystyle \displaystyle \int \frac {dx}{x^2-a^2}=\frac {1}{2a} \ln \big{|} \frac {x-a}{x+a}\big{|} + C $
$\displaystyle \displaystyle \int \frac{dx}{\sqrt{x^2 \pm a^2}}=\ln{\big{|}x+\sqrt{x^2\pm a^2\big{|}}+C$
$\displaystyle \displaystyle \int a^x \; dx = \frac {a^x}{\ln {a}} +C $
and so on we use like 30 maybe "basic" integrals, like table integrals which u can use without calculating them and with some basic operations, and substitutions (and a lot of practice) u can solve probably (for sure) any integral that u can get to do in your school
Let $\displaystyle \displaystyle u = 2-x$ then $\displaystyle \displaystyle \frac{du}{dx} = -1 \Rightarrow {dx} = -{du} \; \therefore \int(2-x)^6\;{dx} = -\int u^6\;{du} = -\frac{1}{7}u^7+k = -\frac{1}{7}(2-x)^7+k
$
After some practice, you will just see by inspection:
$\displaystyle \displaystyle \int(ax+b)^n\;{dx} = \frac{1}{a(n+1)}(ax+b)^{n+1}+k$
Since by letting $\displaystyle u = ax+b$, we have $\displaystyle \displaystyle \dfrac{du}{dx} = a \Rightarrow {dx} = \dfrac{du}{a}$ thus $\displaystyle \displaystyle \int(ax+b)^n\;{dx} = \frac{1}{a}\int u^n\;{du} = \frac{1}{a(n+1)}u^{n+1}+k = \frac{1}{a(n+1)}(ax+b)^{n+1}+k$
(If $\displaystyle a = -1$, and $\displaystyle b = 2$, we have the above problem).
if u substitute $\displaystyle u = 2-x $ and $\displaystyle du=-dx$
aou will get:
$\displaystyle \displaystyle \int \frac {dx}{(2-x)^6} = -\int \frac {du}{u^6} = -\frac {1}{5(2-x)^5} + C $
P.S. sorry I didn't see that is just (2-x)^6 and sorry TheCoffeeMachine didn't see your post
For your original question, you have been told that this integral is $\displaystyle \arctan{x} + C$.
The way they get this result is through trigonometric substitution. Trigonometric substitutions are handy to know, because through the use of the trigonometric identities, you can usually simplify the problem...
Let $\displaystyle x = \tan{\theta}$ so that $\displaystyle dx = \sec^2{\theta}\,d\theta$. Note that this means $\displaystyle \theta = \arctan{x}$.
Then
$\displaystyle \int{\frac{1}{1 + x}\,dx} = \int{\frac{1}{1 + \tan^2{\theta}}\,\sec^2{\theta}\,d\theta}$
$\displaystyle = \int{\frac{\sec^2{\theta}}{1 + \tan^2{\theta}}\,d\theta}$
$\displaystyle = \int{\frac{\sec^2{\theta}}{\sec^2{\theta}}\,d\thet a}$
$\displaystyle = \int{1\,d\theta}$
$\displaystyle = \theta + C$
$\displaystyle = \arctan{x} + C$.