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**Formula for integration by parts:**

[tex]\int f(x)dx = \int u dv = uv - \int v du[/tex]

**Use integration by parts to find the following integrals:**

a) [tex]\int x e^{1-x} dx [/tex]

b) [tex]\int_1^4 \frac {ln \sqrt x} {\sqrt x} dx[/tex]

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

d) [tex]\int x^3 \sqrt{3x^2+2} dx[/tex]

**Answers in back of the book:**

a) [tex]-e^{1-x}(x+1)+C[/tex]

b) 4 ln 2 - 2

c) [tex]\frac{74}{4}[/tex]

d) [tex]\frac {1}{9}x^2(3x^2+2)^{\frac {3}{2}}-\frac{2}{135}(3x^2+2)^{\frac {5}{2}} +C[/tex]

**My attempts:**

a) [tex]\int xe^{(1-x)} dx[/tex]

[tex]u = x[/tex]

[tex]dv = e^{1-x} dx[/tex]

[tex]du = 1 [/tex]

[tex]v = -e^{1-x}[/tex]

which gives:

[tex]-xe^{1-x} - \int -e^{1-x}(1)[/tex]

[tex]-xe^{1-x} - e^{1-x}[/tex]

[tex]-e^{1-x}(x+1) +C[/tex]

GOT IT! Thanks Melawrghk and dx!

b) [tex]\int_1^4 \frac {ln \sqrt x} {\sqrt x} dx[/tex]

[tex]u = ln x^{{1}{2}}[/tex]

[tex]dv = x^{\frac {-1}{2}}[/tex]

[tex]du = \frac { \frac {1}{2}\sqrt x}{\sqrt x} = \frac {1}{2x}[/tex]

[tex]v = 2x^{\frac{1}{2}}[/tex]

which gives:

[tex] ln x^{\frac {1}{2}} 2x^{\frac{1}{2}} - \int 2x^{\frac{1}{2}} \frac {1}{2x}[/tex]

[tex] ln x^{\frac {1}{2}} 2x^{\frac{1}{2}} - \int \frac{x^{\frac {1}{2}}}{x}[/tex]

[tex] ln x^{\frac {1}{2}} 2x^{\frac{1}{2}} - \int x^{\frac {-1}{2}}[/tex]

[tex] ln x^{\frac {1}{2}} 2x^{\frac{1}{2}} - 2x^{\frac{1}{2}}[/tex]

[tex] 2x^{\frac{1}{2}}(ln x^{\frac {1}{2}}-1)[/tex]

Plug in 4 and 1:

[tex]4(ln2-1)-2(ln1-1)[/tex]

[tex] = 4ln2-4+2[/tex]

[tex] = 4ln2-2[/tex]

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.

Thanks in advance!

EDIT: Got

*a*correct now, thanks to Melawrghk and dx.

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