Is This Integration Method Correct for Exponential and Polynomial Functions?

[courtrigrad]

\int 2e^{2x} dx Ok so I know that \int e^{u} du = e^{u} + C. So we have 2\int e^{2x} dx. Now u = 2x, du = 2dx. So dx = \frac{1}{2} du. So now we have \int e^{2x} du = e^{2x} + C Was my method correct?

\int e^{2x} dx Ok for this u = 2x, du = 2dx. So dx = \frac{1}{2} du. So we have \frac{1}{2}\int e^{2x} du = \frac{1}{2} e^{2x} + C. Is this correct?

\int 5x^{3}e^{x^{2}} dx So u = x^{2} du = 2xdx.

 

[MathStudent]

first one yes,, can be verified by taking derivative of solution and seeing if you get back the integrand

second one is good too, just a constant multiple of first.

for third use integration by parts (hint: let u = x^2 and dv = x e^{x^2})

 

[courtrigrad]

Ok so \int 5x^{3}e^{x^{2}} dx so u = x^2 which means du = 2xdx. Does this mean that \frac{1}{2}\int e^{x^{2}} (2xdx) = \frac{1}{2} e^{x^{2}} + C ?

 

[MathStudent]

the formula for integration by parts is given by

\int u dv = uv - \int v du

so for this case letting
u = x^2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)
du = 2x dx \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)

and
dv = x e^{x^2}dx \ \ \ \ \ \ \ \ \ \ \ \ \ (3)
\int dv = \int x e^{x^2}dx \ \ \ \ \ \ (4)

making the substitution w = x^2 in (4) and solving for v you get
v = \frac{1}{2} e^{x^2}

plugging it all into the equation for separation of variables then

\int 5 x^3 e^{x^2} dx = 5[\frac{1}{2} x^2 e^{x^2} - \int 2x \frac{1}{2} e^{x^2} dx]

try to do the rest

 

[courtrigrad]

however we are supposed to do substitution not parts

also is:

\int \frac{1}{x+1} dx = \Ln(x+1) + C?
\int \frac{1}{3-2x} dx = \frac{-1}{2} \Ln(3-2x) + C
\int \frac{e^{2x} + e^{x} + 1}{e^{x}} = e^{x} -e^{-x} + x + 2C?

 

[dextercioby]

You can't do it by substitution only.U have to apply part integration as well...

Daniel.

 

[MathStudent]

however we are supposed to do substitution not parts

also is:

\int \frac{1}{x+1} dx = \ln(x+1) + C?
\int \frac{1}{3-2x} dx = \frac{-1}{2} \ln(3-2x) + C
\int \frac{e^{2x} + e^{x} + 1}{e^{x}} = e^{x} -e^{-x} + x + 2C?

note: improper format for natural log in latex

hint: try taking the derivative of your solutions, and see if they match the integrand

 

[dextercioby]

What happened to the logarithms...?

Is that
\int \frac{dx}{xx} =\int \frac{dx}{x^{2}} [/itex]<br /> <br /> ?<br /> <br /> Daniel.

 

[dextercioby]

HINT:d(\ln x)=\frac{dx}{x}

Daniel.

 

[courtrigrad]

actually it is \int \frac{1}{x\ln x} Ok so \int \frac{\ln x}{x} dx. So u = \ln x and du = \frac{1}{x} dx. That means dx = xdu. Is this correct?

Also for \int \frac{e^{2x} + e^{x} + 1}{e^{x}} = e^{x} -e^{-x} + x + 2C I just divided through. Is this correct?

Thanks

 

[dextercioby]

If u put the "dx" in the integral,then yes...

Daniel.

 

[dextercioby]

Are u talking about the same integral??

\int \frac{dx}{x\ln x} \neq \int -\ln x \frac{dx}{x}

Daniel.

 

[courtrigrad]

Ok so for \int \frac{1}{x\ln x} should I rewrite or just make u = \ln x?

Thanks

 

[dextercioby]

1.Put the "dx"...
2.Yes,make that substitution.

Daniel.

 

[courtrigrad]

ok so i got du = \frac{1}{x} dx dx = xdu. So now do I bring the x out in the front?I know the answer is \ln(\ln(x)))

thanks

 

[Kamataat]

no need to do dx=xdu. just substitute straight from \frac{1}{x}dx=du.

edit: add a +C to ln(lnx).

- kamataat