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

 

[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