Question Video: Using Partial Fractions to Evaluate an Indefinite Integral | Nagwa Question Video: Using Partial Fractions to Evaluate an Indefinite Integral | Nagwa

Question Video: Using Partial Fractions to Evaluate an Indefinite Integral Mathematics

Use partial fractions to evaluate the integral ∫ 4/(π‘₯Β² βˆ’ 5π‘₯ βˆ’ 14) dπ‘₯.

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Video Transcript

Use partial fractions to evaluate the integral the integral of four divided by π‘₯ squared minus five π‘₯ minus 14 with respect to π‘₯.

The question wants us to evaluate our integral, and we can see our integrand is a rational function. We’re told that we should use partial fractions to help us evaluate this integral. So we want to use partial fractions on our integrand. To do this, we need to factor our denominator. And we can see that 14 is seven times two. This helps us factor our quadratic. We can see that negative seven times two is negative 14, and negative seven plus two is equal to negative five. So π‘₯ squared minus five π‘₯ minus 14 is equal to π‘₯ minus seven times π‘₯ plus two.

So our denominator is two unique factors. Therefore, by using partial fractions, we can write four divided by π‘₯ minus seven times π‘₯ plus two as 𝐴 divided by π‘₯ minus seven plus 𝐡 divided by π‘₯ plus two for some constants 𝐴 and 𝐡. To help us find the values of 𝐴 and 𝐡, we’re going to multiply both sides of this equation by π‘₯ minus seven times π‘₯ plus two. If we do this and then simplify, we get four is equivalent to 𝐴 times π‘₯ plus two plus 𝐡 times π‘₯ minus seven. And it’s worth reiterating we use equivalents instead of equals because these are true for all values of π‘₯.

And because this is true for all values of π‘₯, we can remove our variable 𝐴 by substituting π‘₯ is equal to negative two. So substituting π‘₯ is equal to negative two, we get four is equal to 𝐴 times negative two plus two plus 𝐡 times negative two minus seven. Of course, negative two plus two is equal to zero, and negative two minus seven is equal to negative nine. So we get four is equal to negative nine 𝐡. And if we divide both sides of this equation by negative nine, we get negative four over nine is equal to 𝐡.

We can do the same to eliminate our variable 𝐡. We’ll substitute π‘₯ is equal to seven. Doing this, we get four is equal to 𝐴 times seven plus two plus 𝐡 times seven minus seven. This time, seven plus two is equal to nine, and seven minus seven is equal to zero. So we get four is equal to nine 𝐴. And then we just divide through by nine. This gives us four over nine is equal to 𝐴. So by using partial fractions, we’ve shown that we can rewrite our integrand, four divided by π‘₯ minus seven times π‘₯ plus two, as 𝐴 divided by π‘₯ minus seven. And we’ve shown that 𝐴 is equal to four-ninths. Then, we want to add 𝐡 divided by π‘₯ plus two. And we’ve shown that 𝐡 is equal to negative four-ninths. So we can now use this to rewrites our integral.

Doing this, we get the integral of four over nine times π‘₯ minus seven minus four divided by nine times π‘₯ plus two with respect to π‘₯. And now we could see each term in our integrand is of a form which we can integrate. We know for any constant 𝑑, the integral of one divided by π‘₯ plus 𝑑 with respect to π‘₯ is equal to the natural logarithm of the absolute value of π‘₯ plus 𝑑 plus a constant of integration 𝐢. And to write both of our terms in this form, we can just take out a constant factor of four over nine. This gives us four over nine times the integral of one divided by π‘₯ minus seven minus one divided by π‘₯ plus two with respect to π‘₯.

So to integrate our first term, we use our integral rule with 𝑑 equal to negative seven. And to integrate our second term, we use our integral rule with 𝑑 equal to two. This gives us four-ninths times the natural logarithm of the absolute value of π‘₯ minus seven minus the natural logarithm of the absolute value of π‘₯ plus two. And we’ll combine both of these constants of integration into a new one we’ll call 𝐢. And we could leave our answer like this. However, we can distribute four-ninths over our constant 𝐢 inside of our parentheses.

This gives us the following expression. And remember, 𝐢 is a constant. So, 𝐢 times four over nine is also a constant. So we’ll write this as a new constant we’ll call 𝐾. And this gives us our final answer. Therefore, by using partial fractions, we’ve shown the integral of four divided by π‘₯ squared minus five π‘₯ minus 14 with respect to π‘₯ is equal to four over nine times the natural logarithm of the absolute value of π‘₯ minus seven minus the natural logarithm of the absolute value of π‘₯ plus two plus our constant of integration 𝐾.

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