Answer
453k+ 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
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)