Integration - Partial fractions (1 Viewer)

Aysce · Feb 11, 2012

I have found out the values for the fraction so here it is:

$\frac{x}{(x+3)(x-1)}=\frac{3}{4(x+3)} + \frac{1}{4(x-1)}$

Integrate $\frac{x}{(x+3)(x-1)}$ dx

I am partially (lame pun >.>) confused with integrating $\frac{3}{4(x+3)}$ because I don't know what value to multiply the fraction by when using the integration log rule.

nightweaver066 · Feb 11, 2012

Take out the constant.

$\int \frac{3}{4(x + 3)} dx$

$= \frac{3}{4} \int \frac{1}{x + 3} dx$

$= \frac{3}{4}ln|x + 3| + c$

Aysce · Feb 11, 2012

nightweaver066 said:
Take out the constant.

$\int \frac{3}{4(x + 3)} dx$

$= \frac{3}{4} \int \frac{1}{x + 3} dx$

$= \frac{3}{4}ln|x + 3| + c$

Answer says 3/2 ? Probably typo

Alkanes · Feb 11, 2012

nightweaver is correct.

Hezz96 · Oct 10, 2013

Aysce said:
Answer says 3/2 ? Probably typo

Either a typo or you have got the wrong values.

Integration - Partial fractions (1 Viewer)

Aysce

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nightweaver066

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Aysce

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Alkanes

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Hezz96

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