How do you integrate $\int \frac{2 x - 3}{x^{2} - 3 x + 5}$ $int (2x-3)/(x^2-3x+5)$ using substitution?

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How do you integrate $\int \frac{2 x - 3}{x^{2} - 3 x + 5}$ $int (2x-3)/(x^2-3x+5)$ using substitution?

1 Answer

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Answer 1 · 2017-02-17T14:14:02

$\int \frac{2 x - 3}{x^{2} - 3 x + 5} d x = ln (x^{2} - 3 x + 5) + C$ $int (2x-3)/(x^2-3x+5)dx = ln (x^2-3x+5) + C$

Explanation:

Note that:

$\frac{d}{d x} (x^{2} - 3 x + 5) = 2 x - 3$ $d/dx (x^2-3x+5) = 2x-3$

So if we substitute:

$t = x^{2} - 3 x + 5$ $t = x^2-3x+5$

$d t = (2 x - 3) d x$ $dt = (2x-3)dx$

we have:

$\int \frac{2 x - 3}{x^{2} - 3 x + 5} d x = \int \frac{d t}{t} = ln | t | + C = ln (x^{2} - 3 x + 5) + C$ $int (2x-3)/(x^2-3x+5)dx = int (dt)/t = ln abs(t) +C = ln (x^2-3x+5) + C$

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How do you integrate $\int \frac{2 x - 3}{x^{2} - 3 x + 5}$ $int (2x-3)/(x^2-3x+5)$ using substitution?

1 Answer

Explanation:

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How do you integrate ∫2x−3x2−3x+5int (2x-3)/(x^2-3x+5) using substitution?

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How do you integrate $\int \frac{2 x - 3}{x^{2} - 3 x + 5}$ $int (2x-3)/(x^2-3x+5)$ using substitution?