Question

# Solve: $\int{\dfrac{{{x}^{2}}+5x+2}{x+2}}$

Hint: So we have to integrate$\int{\dfrac{{{x}^{2}}+5x+2}{x+2}}$. Then for that substitute$x+2=t$and then simplify it. Use the sum rule and integrate it. You will get the answer.

So we have to integrate$\int{\dfrac{{{x}^{2}}+5x+2}{x+2}}dx$.
Integration is the reverse of differentiation.
However:
If $y=2x+3,\dfrac{dy}{dx}=2$
If $y=2x+5,\dfrac{dy}{dx}=2$
If $y=2x,\dfrac{dy}{dx}=2$
So the integral of $2$ can be$2x+3,2x+5,2x$etc.
For this reason, when we integrate, we have to add a constant. So the integral of$2$ is$2x+c$, where $c$ is a constant.
An "S" shaped symbol is used to mean the integral of, and$dx$is written at the end of the terms to be integrated, meaning "with respect to $x$". This is the same "$dx$" that appears in$\dfrac{dy}{dx}$.

We have to use the substitution method.
There are occasions when it is possible to perform an apparently difficult piece of integration by first making a substitution. This has the effect of changing the variable and the integrand.
So now let us take an example such that$\int{{{(x+1)}^{4}}}dx$,
So for the above, we know we have solved many integrations like$\int{{{u}^{4}}du}$.
So we can see that instead of$x+1$there is$u$.
So$x+1=u$, differentiating we get $dx=du$. So, our integral becomes :
$\int{{{(x+1)}^{4}}}dx=\int{{{u}^{4}}du}=\dfrac{{{u}^{5}}}{5}+c$
So substituting we get,
$=\dfrac{{{(x+1)}^{5}}}{5}+c$
So like this we have to substitute for this sum as well.
So we have to integrate the above integral$\int{\dfrac{{{x}^{2}}+5x+2}{x+2}}dx$.
So now substituting$x+2=t$.
So differentiating we get,
$dx=du$
And $x=t-2$,
So substituting above we get,
$\int{\dfrac{{{x}^{2}}+5x+2}{x+2}}dx=\int{\dfrac{{{(t-2)}^{2}}+5(t-2)+2}{t}}du$
So simplifying we get,
$=\int{\dfrac{{{(t-2)}^{2}}+5(t-2)+2}{t}}du=\int{\dfrac{{{t}^{2}}-4t+4+5t-10+2}{t}}du=\int{\dfrac{{{t}^{2}}+t-4}{t}}du$
So now splitting the terms we get,
$=\left( \int{\dfrac{{{t}^{2}}}{t}+\dfrac{t}{t}}-\dfrac{4}{t} \right)dt$
Next, weâ€™ll simplify and apply the sum rule.
Sum rule is$\int{(a+b)}dx=\int{adx+}\int{bdx}$.
So, applying it and simplifying further, we get,
$=\left( \int{\dfrac{{{t}^{2}}}{t}+\dfrac{t}{t}}-\dfrac{4}{t} \right)dt=\int{tdt+\int{dt-\int{\dfrac{4}{t}dt}}}$
Now, letâ€™s try applying integration.
We know, that$\int{{{p}^{n}}dp=\dfrac{{{p}^{n+1}}}{n+1}}+c$and$\int{\dfrac{1}{p}dp=\log p+c}$
So applying these properties, we get,
$=\dfrac{{{t}^{2}}}{2}+t-4\log t+c$
So now substituting the value, we get,
$=\dfrac{{{(x+2)}^{2}}}{2}+(x+2)-4\log (x+2)+c$
So we get the final answer $\int{\dfrac{{{x}^{2}}+5x+2}{x+2}}=\dfrac{{{(x+2)}^{2}}}{2}+(x+2)-4\log (x+2)+c$

Note: You should know the basic things of integration. So here substitution is important. It depends on you what you are substituting. You should know$\int{{{p}^{n}}dp=\dfrac{{{p}^{n+1}}}{n+1}}+c$and$\int{{{p}^{n}}dp=\dfrac{{{p}^{n+1}}}{n+1}}+c$. Also, you must know the rules of integration. Avoid silly mistakes because silly mistakes change the whole problem.