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

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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.

Complete step-by-step 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, 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,

$=\left( \int{\dfrac{{{t}^{2}}}{t}+\dfrac{t}{t}}-\dfrac{4}{t} \right)dt=\int{tdt+\int{dt-\int{\dfrac{4}{t}dt}}}$

Now, let’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.

Complete step-by-step 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, 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,

$=\left( \int{\dfrac{{{t}^{2}}}{t}+\dfrac{t}{t}}-\dfrac{4}{t} \right)dt=\int{tdt+\int{dt-\int{\dfrac{4}{t}dt}}}$

Now, let’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.

Last updated date: 24th Sep 2023

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