What fraction is equivalent to 4 over negative 9

Evaluate the following integral (Calculus 2) Please provide step by step explanation!

Step2247 [10]

Answer:

$\displaystyle \int \dfrac{2}{x^2+2x+1}\:\:\text{d}x=-\dfrac{2}{x+1}+\text{C}$

Step-by-step explanation:

Fundamental Theorem of Calculus

$\displaystyle \int \text{f}(x)\:\text{d}x=\text{F}(x)+\text{C} \iff \text{f}(x)=\dfrac{\text{d}}{\text{d}x}(\text{F}(x))$

If differentiating takes you from one function to another, then integrating the second function will take you back to the first with a constant of integration.

Given integral:

$\displaystyle \int \dfrac{2}{x^2+2x+1}\:\:\text{d}x$

Factor the denominator:

$\begin{aligned}\implies x^2+2x+1 & = x^2+x+x+1\\& = x(x+1)+1(x+1)\\& = (x+1)(x+1)\\& = (x+1)^2\end{aligned}$

$\implies \displaystyle \int \dfrac{2}{x^2+2x+1}\:\:\text{d}x=\int \dfrac{2}{(x+1)^2}\:\:\text{d}x$

$\textsf{Apply exponent rule} \quad \dfrac{1}{a^n}=a^{-n}$

$\implies \displaystyle \int \dfrac{2}{x^2+2x+1}\:\:\text{d}x=\int 2(x+1)^{-2}\:\:\text{d}x$

$\boxed{\begin{minipage}{4 cm}\underline{Integrating $ax^n$}\\\\$\displaystyle \int ax^n\:\text{d}x=\dfrac{ax^{n+1}}{n+1}+\text{C}$\end{minipage}}$

Use Integration by Substitution:

$\textsf{Let }u=(x+1) \implies \dfrac{\text{d}u}{\text{d}x}=1 \implies \text{d}x=\text{d}u}$

Therefore:

$\begin{aligned}\displaystyle \int \dfrac{2}{x^2+2x+1}\:\:\text{d}x & = \int 2(x+1)^{-2}\:\:\text{d}x\\\\& = \int 2u^{-2}\:\:\text{d}u\\\\& = \dfrac{2}{-1}u^{-2+1}+\text{C}\\\\& = -2u^{-1}+\text{C}\\\\& = -\dfrac{2}{u}+\text{C}\\\\& = -\dfrac{2}{x+1}+\text{C}\end{aligned}$