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

Last updated date: 28th Mar 2023

•

Total views: 308.1k

•

Views today: 3.85k

Answer

Verified

308.1k+ views

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.

Recently Updated Pages

If a spring has a period T and is cut into the n equal class 11 physics CBSE

A planet moves around the sun in nearly circular orbit class 11 physics CBSE

In any triangle AB2 BC4 CA3 and D is the midpoint of class 11 maths JEE_Main

In a Delta ABC 2asin dfracAB+C2 is equal to IIT Screening class 11 maths JEE_Main

If in aDelta ABCangle A 45circ angle C 60circ then class 11 maths JEE_Main

If in a triangle rmABC side a sqrt 3 + 1rmcm and angle class 11 maths JEE_Main

Trending doubts

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE