21.04.2022

Integral of x/(x + 1)²(x+2)²

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Step-by-step answer

09.07.2023, solved by verified expert
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I suppose you mean the integral

Integral of x/(x + 1)²(x+2)², №18010190, 21.04.2022 13:21

Break up the integrand into partial fractions:

Integral of x/(x + 1)²(x+2)², №18010190, 21.04.2022 13:21

Integral of x/(x + 1)²(x+2)², №18010190, 21.04.2022 13:21

Integral of x/(x + 1)²(x+2)², №18010190, 21.04.2022 13:21

Integral of x/(x + 1)²(x+2)², №18010190, 21.04.2022 13:21

Solve for the coefficients:

Integral of x/(x + 1)²(x+2)², №18010190, 21.04.2022 13:21

So we have

Integral of x/(x + 1)²(x+2)², №18010190, 21.04.2022 13:21

Integral of x/(x + 1)²(x+2)², №18010190, 21.04.2022 13:21

Integral of x/(x + 1)²(x+2)², №18010190, 21.04.2022 13:21

Integral of x/(x + 1)²(x+2)², №18010190, 21.04.2022 13:21

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Faq

Mathematics
Step-by-step answer
P Answered by Specialist

I suppose you mean the integral

\displaystyle \int \frac{x}{(x+1)^2 (x+2)^2} \, dx

Break up the integrand into partial fractions:

\dfrac{x}{(x+1)^2 (x+2)^2} = \dfrac{a}{x+1} + \dfrac{b}{(x+1)^2} + \dfrac{c}{x+2} + \dfrac{d}{(x+2)^2}

\dfrac{x}{(x+1)^2 (x+2)^2} \\ = \dfrac{a(x+1)(x+2)^2 + b(x+2)^2 + c(x+1)^2(x+2) + d(x+1)^2}{(x+1)^2 (x+2)^2}

\dfrac{x}{(x+1)^2 (x+2)^2} \\ = \dfrac{(a+c)x^3+(5a+b+4c+d)x^2+(8a+4b+5c+2d)x + 4a+4b+2c+d}{(x+1)^2(x+2)^2}

x = (a+c)x^3+(5a+b+4c+d)x^2+(8a+4b+5c+2d)x + 4a+4b+2c+d

Solve for the coefficients:

\begin{cases}a+c=0 \\ 5a+b+4c+d=0 \\ 8a+4b+5c+2d=1 \\ 4a+4b+2c+d=0\end{cases} \implies a=3,b=-1,c=-3,d=-2

So we have

\displaystyle \int \frac{x}{(x+1)^2 (x+2)^2} \, dx= \int \left(\frac3{x+1} - \frac1{(x+1)^2} - \frac3{x+2} - \frac2{(x+2)^2}\right) \, dx

\displaystyle \int \frac{x}{(x+1)^2 (x+2)^2} \, dx= 3\ln|x+1| + \frac1{x+1} - 3\ln|x+2| + \frac2{x+2} + C

\displaystyle \int \frac{x}{(x+1)^2 (x+2)^2} \, dx= \ln|x+1|^3 - \ln|x+2|^3 + \frac{3x+4}{(x+1)(x+2)} + C

\displaystyle \int \frac{x}{(x+1)^2 (x+2)^2} \, dx= \boxed{\ln\left|\frac{x+1}{x+2}\right|^3 + \frac{3x+4}{(x+1)(x+2)} + C}

Mathematics
Step-by-step answer
P Answered by PhD

(x+y)^2\,\mathrm dx+(2xy+x^2-2)\,\mathrm dy=0

Suppose the ODE has a solution of the form F(x,y)=C, with total differential

\dfrac{\partial F}{\partial x}\,\mathrm dx+\dfrac{\partial F}{\partial y}\,\mathrm dy=0

This ODE is exact if the mixed partial derivatives are equal, i.e.

\dfrac{\partial^2F}{\partial y\partial x}=\dfrac{\partial^2F}{\partial x\partial y}

We have

\dfrac{\partial F}{\partial x}=(x+y)^2\implies\dfrac{\partial^2F}{\partial y\partial x}=2(x+y)

\dfrac{\partial F}{\partial y}=2xy+x^2-2\implies\dfrac{\partial^2F}{\partial x\partial y}=2y+2x=2(x+y)

so the ODE is indeed exact.

Integrating both sides of

\dfrac{\partial F}{\partial x}=(x+y)^2

with respect to x gives

F(x,y)=\dfrac{(x+y)^3}3+g(y)

Differentiating both sides with respect to y gives

\dfrac{\partial F}{\partial y}=2xy+x^2-2=(x+y)^2+\dfrac{\mathrm dg}{\mathrm dy}

\implies x^2+2xy-2=x^2+2xy+y^2+\dfrac{\mathrm dg}{\mathrm dy}

\implies\dfrac{\mathrm dg}{\mathrm dy}=-y^2-2

\implies g(y)=-\dfrac{y^3}3-2y+C

\implies F(x,y)=\dfrac{(x+y)^3}3-\dfrac{y^3}3-2y+C

so the general solution to the ODE is

F(x,y)=\dfrac{(x+y)^3}3-\dfrac{y^3}3-2y=C

Given that y(1)=1, we find

\dfrac{(1+1)^3}3-\dfrac{1^3}3-2=C\implies C=\dfrac13

so that the solution to the IVP is

F(x,y)=\dfrac{(x+y)^3}3-\dfrac{y^3}3-2y=\dfrac13

\implies\boxed{(x+y)^3-y^3-6y=1}

Mathematics
Step-by-step answer
P Answered by PhD

(x+y)^2\,\mathrm dx+(2xy+x^2-2)\,\mathrm dy=0

Suppose the ODE has a solution of the form F(x,y)=C, with total differential

\dfrac{\partial F}{\partial x}\,\mathrm dx+\dfrac{\partial F}{\partial y}\,\mathrm dy=0

This ODE is exact if the mixed partial derivatives are equal, i.e.

\dfrac{\partial^2F}{\partial y\partial x}=\dfrac{\partial^2F}{\partial x\partial y}

We have

\dfrac{\partial F}{\partial x}=(x+y)^2\implies\dfrac{\partial^2F}{\partial y\partial x}=2(x+y)

\dfrac{\partial F}{\partial y}=2xy+x^2-2\implies\dfrac{\partial^2F}{\partial x\partial y}=2y+2x=2(x+y)

so the ODE is indeed exact.

Integrating both sides of

\dfrac{\partial F}{\partial x}=(x+y)^2

with respect to x gives

F(x,y)=\dfrac{(x+y)^3}3+g(y)

Differentiating both sides with respect to y gives

\dfrac{\partial F}{\partial y}=2xy+x^2-2=(x+y)^2+\dfrac{\mathrm dg}{\mathrm dy}

\implies x^2+2xy-2=x^2+2xy+y^2+\dfrac{\mathrm dg}{\mathrm dy}

\implies\dfrac{\mathrm dg}{\mathrm dy}=-y^2-2

\implies g(y)=-\dfrac{y^3}3-2y+C

\implies F(x,y)=\dfrac{(x+y)^3}3-\dfrac{y^3}3-2y+C

so the general solution to the ODE is

F(x,y)=\dfrac{(x+y)^3}3-\dfrac{y^3}3-2y=C

Given that y(1)=1, we find

\dfrac{(1+1)^3}3-\dfrac{1^3}3-2=C\implies C=\dfrac13

so that the solution to the IVP is

F(x,y)=\dfrac{(x+y)^3}3-\dfrac{y^3}3-2y=\dfrac13

\implies\boxed{(x+y)^3-y^3-6y=1}

Mathematics
Step-by-step answer
P Answered by Specialist

F'(2) = 3

\int\limits^x_0 {\sqrt{t^3+1}} \, dt

Using Fundamental Theorem of Calculus:

Since x = 2, you now have:

\int\limits^2_0 {\sqrt{t^3+1} } \, dt

so F'(x) = \sqrt{t^3+1}

--> F'(2) = \sqrt{2^3+1} = \sqrt{9} = 3

so F'(2) = 3

Mathematics
Step-by-step answer
P Answered by Specialist

F'(2) = 3

\int\limits^x_0 {\sqrt{t^3+1}} \, dt

Using Fundamental Theorem of Calculus:

Since x = 2, you now have:

\int\limits^2_0 {\sqrt{t^3+1} } \, dt

so F'(x) = \sqrt{t^3+1}

--> F'(2) = \sqrt{2^3+1} = \sqrt{9} = 3

so F'(2) = 3

Mathematics
Step-by-step answer
P Answered by Master

Step-by-step explanation:

Given is the function

xy =3

Use product rule to find derivative

xy'+y =0 OR y'=\frac{-y}{x}

Arc length = \int\limits^1_6 {\sqrt{1+\frac{y^2}{x^2} } } \, dx \\=\int\limits^1_6 {\sqrt{\frac{x^2+y^2}{x^2} } } \, dx \\\\

Substitute y = 3/x

WE get

arc length = \int\limits^1_6 {\sqrt{1+\frac{9}{x^4} } } \, dx \\\\=\int\limits^1_6 {\sqrt{\frac{x^4+9}{x^4} } } \, dx \\

Using simpson rule for 1/3 we have

x(x^4+9)^1/2f(x)2 times or 4 times  

13.162277663.162277663.16227766  

1.53.751.6666666676.666666667  

251.25  

2.56.9327123121.109233974.43693588  

39.4868329811.054092553  

3.512.611998261.0295508784.118203512  

416.27882061.0174262872.034852575  

4.520.47101611.0109143754.0436575  

525.179356621.0071742652.01434853  

5.530.398396341.0049056644.019622656  

636.124783741.0034662152.00693243  

6.542.356374961.0025177514.010071002  

749.091750831.0018724662.003744932  

7.556.329943191.0014212124.005684849  

864.070273921.001098032.00219606  

8.572.312256911.0008616874.00344675  

981.055536521.0006856362.001371272  

9.590.299847731.0005523294.002209318  

10100.04498991.0004498992.000899798  

10.5110.29080881.0003701484.00148059  

11121.03718441.0003073092.000614618  

11.5132.28402211.0002572564.001029023  

12144.03124661.000216994.82148E-05  

   

 52.270413638.711735605

Answer is 8.7117

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