# Integration by Parts: Find Integrals | 65 Characters

• aleao
My bad!Yes, the best way to proceed is to split it up like that and use the substitution u=3x^2+2. Sorry again for the goof up.In summary, the formula for integration by parts is \int f(x)dx = \int u dv = uv - \int v du. Using this method, we can solve for the following integrals:a) \int x e^{1-x} dx = -e^{1-x}(x+1)+Cb) \int_1^4 \frac {ln \sqrt x} {\sqrt x} dx = 4ln2-2c) \int_{-2}^1 (2x+1)(

#### aleao

Formula for integration by parts:

$$\int f(x)dx = \int u dv = uv - \int v du$$

Use integration by parts to find the following integrals:

a) $$\int x e^{1-x} dx$$

b) $$\int_1^4 \frac {ln \sqrt x} {\sqrt x} dx$$

c) $$\int_{-2}^1 (2x+1)(x+3)^{3/2} dx$$

d) $$\int x^3 \sqrt{3x^2+2} dx$$

Answers in back of the book:

a) $$-e^{1-x}(x+1)+C$$

b) 4 ln 2 - 2

c) $$\frac{74}{4}$$

d) $$\frac {1}{9}x^2(3x^2+2)^{\frac {3}{2}}-\frac{2}{135}(3x^2+2)^{\frac {5}{2}} +C$$

My attempts:

a) $$\int xe^{(1-x)} dx$$

$$u = x$$

$$dv = e^{1-x} dx$$

$$du = 1$$

$$v = -e^{1-x}$$

which gives:

$$-xe^{1-x} - \int -e^{1-x}(1)$$

$$-xe^{1-x} - e^{1-x}$$

$$-e^{1-x}(x+1) +C$$

GOT IT! Thanks Melawrghk and dx!

b) $$\int_1^4 \frac {ln \sqrt x} {\sqrt x} dx$$

$$u = ln x^{{1}{2}}$$

$$dv = x^{\frac {-1}{2}}$$

$$du = \frac { \frac {1}{2}\sqrt x}{\sqrt x} = \frac {1}{2x}$$

$$v = 2x^{\frac{1}{2}}$$

which gives:

$$ln x^{\frac {1}{2}} 2x^{\frac{1}{2}} - \int 2x^{\frac{1}{2}} \frac {1}{2x}$$

$$ln x^{\frac {1}{2}} 2x^{\frac{1}{2}} - \int \frac{x^{\frac {1}{2}}}{x}$$

$$ln x^{\frac {1}{2}} 2x^{\frac{1}{2}} - \int x^{\frac {-1}{2}}$$

$$ln x^{\frac {1}{2}} 2x^{\frac{1}{2}} - 2x^{\frac{1}{2}}$$

$$2x^{\frac{1}{2}}(ln x^{\frac {1}{2}}-1)$$

Plug in 4 and 1:

$$4(ln2-1)-2(ln1-1)$$

$$= 4ln2-4+2$$

$$= 4ln2-2$$

GOT IT! Thanks again dx!

I'm looking at the other two and I'm blanking.

I've been out of school for three years, and jumped right into business calculus since it's a requirement. So I apologize if this is elementary, but I've been banging my head on this for hours now... and of course the teacher is nowhere to be found.

EDIT: Got a correct now, thanks to Melawrghk and dx.

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For a), switch your u and dv and try again :)

EDIT: and for b), check your derivative of ln(sqrt(x)). sqrt(x) can be written as x^(1/2), so when you take the derivative you'll get (1/2)x^(3/2). Coupled with the square root of x on the bottom it will cancel nicely.

For part (a), v = e1-x would be a better choice.

Is $$\int e^{1-x} = e^{1-x}$$ ?

Edit: I'm only assuming this because

$$\int e^{x} = e^{x}$$

... but now that I think about it...

$$\int e^{1-x} = \frac {1}{1-x} e^{1-x+1} = \frac {1}{1-x} e^{x}$$

right?

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Melawrghk said:
EDIT: and for b), check your derivative of ln(sqrt(x)). sqrt(x) can be written as x^(1/2), so when you take the derivative you'll get (1/2)x^(3/2). Coupled with the square root of x on the bottom it will cancel nicely.

Wouldn't the derivative of x^(1/2) = (1/2)x^(-1/2)?

aleao said:
Is $$\int e^{1-x} = e^{1-x}$$ ?

Edit: I'm only assuming this because

$$\int e^{x} = e^{x}$$

... but now that I think about it...

$$\int e^{1-x} = \frac {1}{1-x} e^{1-x+1} = \frac {1}{1-x} e^{x}$$

right?

Nope. ∫ e1-x dx = -e1-x. Use the substitution t = 1-x to see why.

aleao said:
Wouldn't the derivative of x^(1/2) = (1/2)x^(-1/2)?

Yes, that's correct.

dx said:
Nope. ∫ e1-x dx = -e1-x. Use the substitution t = 1-x to see why.

Aha! Great, now I got part a and edited up there. Thanks :)

For part (b), use v = √x.

dx said:
For part (b), use v = √x.

Could you walk me through this method, please?

I'm taught to look at the equation and assign u and dv. Then derive u to get du and integrate dv to get v.

The book says:

Step 1: Choose functions u and v so that f(x)dx = u dv. Try to pick u so that du is simpler than u and a dv that is easy to integrate.

Step 2: Organize the computation... (they put it in a neat box :tongue:)

Step 3: Complete the integration by finding ∫f(x) dx = ∫ u dv = uv - ∫ v du

So I'm not sure how I assign v = √x directly.

Here's the problem again:

$$\int_1^4 \frac {ln \sqrt x} {\sqrt x} dx$$

Hint: If v = √x, then

$$dv = \frac{dx}{2\sqrt{x}}$$

Got part b! Dx, you rock!

OK so for part d I'm running into a seemingly basic problem. I'm not sure how to integrate √(3x2+2)

I'm just fried...

aleao said:
OK so for part d I'm running into a seemingly basic problem. I'm not sure how to integrate √(3x2+2)

I'm just fried...

You have a direct formula for that.

What is the formula? I'm trying to study for tomorrow morning's final, doing these practice questions and papers are everywhere.

http://planetmath.org/encyclopedia/IntegrationFormulas.html [Broken]

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Don't use a formula, splitting it up in x^3 and the root is a poor way to split things. Do it like this instead.

$$x^3 \sqrt{3x^2+2}=x^2*x\sqrt{3x^2+2}$$. Now use partial integration and note that $x\sqrt{3x^2+2}$ is really easy to integrate by substituting u=3x^2+2, or even "guessing" the primitive.

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Oops.. My apologies... I just read the post no 13 and not the original post. So i just thought its $sqrt{3x^2+2}$ and didnt notice the x cube