Questions & Answers
Question
Answers
Related Questions
Solve: $\int{\dfrac{{{x}^{2}}+5x+2}{x+2}}$
Answer
![Verified]()
Verified
Verified
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.
×
Sorry!, This page is not available for now to bookmark.
Like this
Liked
-
Courses-
Crash Courses 2021
-
Olympiads/NTSE/Others
-
Micro Courses
-
Vedantu Super Kids NEW
-
Reading & Spoken English -
FREEMaster Classes
-
PREMIUM1-on-1 Classes
-
Study Material
-
NCERT Solutions
-
Previous Year Papers
-
Mock Tests
-
Sample Papers
-
Reference Book Solutions
-
ICSE Solutions
-
School Syllabus
-
Revision Notes
-
Important Questions
-
Math Formula Sheets
-
-
More
Book your FREE Online counselling session
Share your contact information
Your FREE Online counselling session has been booked!
Vedantu academic counsellor will be calling you shortly for your Online Counselling session.
DONE
Please try again later
OK
NCERT Book Solutions
Reference Book Solutions
Toll Free:
1800-120-456-456
Mobile:
+91 988-660-2456
Toll Free:
1800-120-456-456
Mobile:
+91 988-660-2456
(Mon-Sun: 9am - 11pm IST)
Questions?
Chat with us
