Question

Solve: $\int {\displaystyle \frac{x^2+5x+2}{x+2}}$

Answer 1

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

Complete step-by-step answer:
So we have to integrate$\int {\displaystyle \frac{x^2+5x+2}{x+2}}dx$.
Integration is the reverse of differentiation.
However:
If $y=2x+3,{\displaystyle \frac{dy}{dx}}=2$
If $y=2x+5,{\displaystyle \frac{dy}{dx}}=2$
If $y=2x,{\displaystyle \frac{dy}{dx}}=2$
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${\displaystyle \frac{dy}{dx}}$.

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 :
$\int {(x+1)}^{4}dx=\int u^{4}du={\displaystyle \frac{u^5}{5}}+c$
So substituting we get,
$={\displaystyle \frac{{(x+1)}^5}{5}}+c$
So like this we have to substitute for this sum as well.
So we have to integrate the above integral$\int {\displaystyle \frac{x^2+5x+2}{x+2}}dx$.
So now substituting$x+2=t$.
So differentiating we get,
$dx=du$
And $x=t-2$,
So substituting above we get,
$\int {\displaystyle \frac{x^2+5x+2}{x+2}}dx=\int {\displaystyle \frac{{(t-2)}^2+5(t-2)+2}{t}}du$
So simplifying we get,
$=\int {\displaystyle \frac{{(t-2)}^2+5(t-2)+2}{t}}du=\int {\displaystyle \frac{t^2-4t+4+5t-10+2}{t}}du=\int {\displaystyle \frac{t^2+t-4}{t}}du$
So now splitting the terms we get,
$=(\int {\displaystyle \frac{t^2}{t}}+{\displaystyle \frac{t}{t}}-{\displaystyle \frac{4}{t}})dt$
Next, we’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,
$=(\int {\displaystyle \frac{t^2}{t}}+{\displaystyle \frac{t}{t}}-{\displaystyle \frac{4}{t}})dt=\int tdt+\int dt-\int {\displaystyle \frac{4}{t}}dt$
Now, let’s try applying integration.
We know, that$\int p^{n}dp={\displaystyle \frac{p^{n+1}}{n+1}}+c$and$\int {\displaystyle \frac{1}{p}}dp=\log p+c$
So applying these properties, we get,
$={\displaystyle \frac{t^2}{2}}+t-4\log t+c$
So now substituting the value, we get,
$={\displaystyle \frac{{(x+2)}^2}{2}}+(x+2)-4\log (x+2)+c$
So we get the final answer $\int {\displaystyle \frac{x^2+5x+2}{x+2}}={\displaystyle \frac{{(x+2)}^2}{2}}+(x+2)-4\log (x+2)+c$

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$\int p^{n}dp={\displaystyle \frac{p^{n+1}}{n+1}}+c$and$\int p^{n}dp={\displaystyle \frac{p^{n+1}}{n+1}}+c$. Also, you must know the rules of integration. Avoid silly mistakes because silly mistakes change the whole problem.