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Integrate \int\frac{ 3x}{x^2 - 7x - 30} dx

Question:

Integrate {eq}\int\frac{ 3x}{x^2 - 7x - 30} dx {/eq}

Indefinite Integral:

Assume that {eq}j {/eq} is a single-valued function of {eq}x {/eq} and {eq}J {/eq} is an anti derivative of {eq}j {/eq} such that {eq}\displaystyle j\left( x \right) = J'\left( x \right),\;\forall \;x {/eq}.

Then the indefinite integral of {eq}\displaystyle j {/eq} is defined as {eq}\displaystyle \int j\left( x \right)dx = J\left( x \right) + c {/eq}, where {eq}c {/eq} is an integration constant.

The following rules are relevant to this problem:

1.{eq}{\int {k\left( {f\left( x \right) + r\left( x \right)} \right)dx = k\int {\left( {f\left( x \right)} \right)dx + k\int {\left( {r\left( x \right)} \right)dx} } } } {/eq}

2.{eq}{\int {\frac{1}{x}dx = \ln \left| x \right| + c} } {/eq}

Answer and Explanation:

Given that: {eq}\displaystyle \int {\frac{{3x}}{{{x^2} - 7x - 30}}} dx {/eq}

{eq}\displaystyle \eqalign{ & \int {\frac{{3x}}{{{x^2} - 7x - 30}}} dx \cr & {\text{Let,}} \cr & f\left( x \right) = \frac{{3x}}{{{x^2} - 7x - 30}} \cr & {\text{From factor;}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{3x}}{{{x^2} - 10x + 3x - 30}} \cr & \,\,\,\,\,\,\,\,\,\,\,\, = \frac{{3x}}{{\left( {x - 10} \right)\left( {x + 3} \right)}} \cr & {\text{From partial fraction;}} \cr & \,\,\,\,\,\,\,\,\,\,\,\, = \frac{9}{{13\left( {x + 3} \right)}} + \frac{{30}}{{13\left( {x - 10} \right)}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\frac{9}{{13\left( {x + 3} \right)}} + \frac{{30}}{{13\left( {x - 10} \right)}} = \frac{{3x}}{{\left( {x - 10} \right)\left( {x + 3} \right)}}} \right) \cr & \cr & f\left( x \right) = \frac{1}{{13}}\left( {\frac{9}{{\left( {x + 3} \right)}} + \frac{{30}}{{\left( {x - 10} \right)}}} \right) \cr & \cr & \int {\frac{{3x}}{{{x^2} - 7x - 30}}} dx \cr & \int {\frac{1}{{13}}\left( {\frac{9}{{\left( {x + 3} \right)}} + \frac{{30}}{{\left( {x - 10} \right)}}} \right)} dx \cr & \frac{9}{{13}}\int {\frac{1}{{\left( {x + 3} \right)}}} dx + \frac{{30}}{{13}}\int {\frac{1}{{\left( {x - 10} \right)}}} dx\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\int {k\left( {f\left( x \right) + r\left( x \right)} \right)dx = k\int {\left( {f\left( x \right)} \right)dx + k\int {\left( {r\left( x \right)} \right)dx} } } } \right) \cr & \frac{9}{{13}}\ln \left| {x + 3} \right| + \frac{{30}}{{13}}\ln \left| {x - 10} \right| + c\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\int {\frac{1}{x}dx = \ln \left| x \right| + c} } \right) \cr & \cr & \int {\frac{{3x}}{{{x^2} - 7x - 30}}} dx = \frac{9}{{13}}\ln \left| {x + 3} \right| + \frac{{30}}{{13}}\ln \left| {x - 10} \right| + c \cr} {/eq}


Learn more about this topic:

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Indefinite Integral: Definition, Rules & Examples

from Calculus: Tutoring Solution

Chapter 7 / Lesson 14
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