Integrals best solved by...

**Latszer** · January 19th 2010, 10:34 PM

For each of the following integrals, determine whether integration by substitution or integration by parts is more appropriate. Do not evaluate the integrals.

(a) $\int x^2cos(x^3) dx$
(b) $\int \frac{1}{sqrt(3x+1)} dx$
(c) $\int x^2sin(x) dx$
(d) $\int ln(x) dx$
(e) $\int xsin(x) dx$
(f) $\int \frac{x^2}{1+x^3}dx$
(g) $\int xe^(x^2) dx$

Which of the integrals are more appropriately found using integration by substitution?

Which of the integrals are more appropriately found using integration by parts?

I have b,d, and f, for sub and a,c,e, and g for by parts.

**Jhevon** · January 19th 2010, 10:43 PM

Originally Posted by Latszer

For each of the following integrals, determine whether integration by substitution or integration by parts is more appropriate. Do not evaluate the integrals.

(a) $\int x^2cos(x^3) dx$
(b) $\int \frac{1}{sqrt(3x+1)} dx$
(c) $\int x^2sin(x) dx$
(d) $\int ln(x) dx$
(e) $\int xsin(x) dx$
(f) $\int \frac{x^2}{1+x^3}dx$
(g) $\int xe^{x^2} dx$

Which of the integrals are more appropriately found using integration by substitution?

Which of the integrals are more appropriately found using integration by parts?

I have b,d, and f, for sub and a,c,e, and g for by parts.

so you just randomly guessed the answers? didn't you even consider checking? explain to me how you would integrate, say, d using substitution.

for this to be more beneficial, you should probably go back over your answers. if you said substitution, say what you would substitute. if you said by parts, say what the parts would be.

**Latszer** · January 19th 2010, 10:59 PM

D would be integration by parts,

u = ln(x)
dv=dx
du=1/x
v=x

$uv-\int du*v dx$
$x*ln(x)-\int x*\frac{1}{x} /dx$
$x*ln(x)-\int 1 dx$
$x*ln(x)-x+c$

Sorry, for being half-assed on my OP, are there any others that I should relook at?

**Jhevon** · January 19th 2010, 11:02 PM

Originally Posted by Latszer

D would be integration by parts,

u = ln(x)
dv=dx
du=1/x
v=x

$uv-\int du*v dx$
$x*ln(x)-\int x*\frac{1}{x} /dx$
$x*ln(x)-\int 1 dx$
$x*ln(x)-x+c$

Sorry, for being half-assed on my OP, are there any others that I should relook at?

hehe, yes, there are others to reconsider. just redo the problem and come up with your new answer set

**Latszer** · January 19th 2010, 11:26 PM

B.
$\int \frac{1}{\sqrt{3x+1}} dx$

Let $u=3x+1$
$\frac{1}{3}du=dx$
$\int \frac{1}{\sqrt{u}}*\frac{1}{3} du$
$\frac{1}{3}\int\frac{1}{\sqrt{u}} du$
$\frac{1}{3}*2u+c$
$\frac{2u}{3}+c$
$\frac{6x+2}{3}$
$2x+\frac{2}{3}$

F
$\int\frac{x^2}{1+x^3}dx$
Let $u=1+x^3$
$3x^2dx=du$
$x^2dx=\frac{1}{3}du$
$\int\frac{1}{1+x^3}*x^2dx$
$\int\frac{1}{u}*\frac{1}{3}du$
$\frac{1}{3}\int\frac{1}{u}du$
$\frac{1}{3}*ln|u|+C$
$\frac{ln|u|}{3}+C$
$\frac{ln|1+x^3|}{3}$

Those look good before I move on?

**Latszer** · January 19th 2010, 11:32 PM

E.

$\int x*sin(x)dx$
u=x
dv=sin x
du=1
v=-cos x

$-x*cos(x)-\int -cos(x)dx$ =-x*cos(x)+sin(x)+C

And from this I can also assume C is also solved from integration by parts.

G.
$x*e^{x^2}dx$
let $u=x^2$
$2xdx=du, so... xdx=\frac{1}{2}du$
$e^u\frac{1}{2}du=\frac{1}{2}\int e^udu$
$\frac{e^u}{2}+C$

Have I proven myself worthy for you to confirm a,b,f, and g are done by substitution and c,d, and e are done with integration by parts?

**Jhevon** · January 19th 2010, 11:36 PM

Originally Posted by Latszer

B.
$\int \frac{1}{\sqrt{3x+1}} dx$

Let $u=3x+1$
$\frac{1}{3}du=dx$
$\int \frac{1}{\sqrt{u}}*\frac{1}{3} du$
$\frac{1}{3}\int\frac{1}{\sqrt{u}} du$
$\frac{1}{3}*2u+c$
$\frac{2u}{3}+c$
$\frac{6x+2}{3}$
$2x+\frac{2}{3}$

correct approach, but you integrated incorrectly. note that $\frac 1{\sqrt u} = u^{-1/2}$

F
$\int\frac{x^2}{1+x^3}dx$
Let $u=1+x^3$
$3x^2dx=du$
$x^2dx=\frac{1}{3}du$
$\int\frac{1}{1+x^3}*x^2dx$
$\int\frac{1}{u}*\frac{1}{3}du$
$\frac{1}{3}\int\frac{1}{u}du$
$\frac{1}{3}*ln|u|+C$
$\frac{ln|u|}{3}+C$
$\frac{ln|1+x^3|}{3}$

Those look good before I move on?

this is fine.

**Jhevon** · January 19th 2010, 11:38 PM

Originally Posted by Latszer

E.

$\int x*sin(x)dx$
u=x
dv=sin x
du=1
v=-cos x

$-x*cos(x)-\int -cos(x)dx$ =-x*cos(x)+sin(x)+C

And from this I can also assume C is also solved from integration by parts.

this is fine. and yes, C is similar to this one. you'd have to use by parts twice.

i am going to leave, so i will leave you with the answers for a and g (that is, the method to use. try to figure it out before you look). you can actually do the integrals for practice.

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