/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 43 Evaluate the following integrals... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 43

Evaluate the following integrals. $$\int \frac{x^{3}}{\left(81-x^{2}\right)^{2}} d x$$

Short Answer

Expert verified
Question: Evaluate the integral: $$\int \frac{x^{3}}{\left(81-x^{2}\right)^{2}} d x$$ Answer: $$-\frac{1}{2} \ln|81 - x^2| + C$$

Step by step solution

01

Choose a suitable substitution

Let's make a substitution for the denominator, which will simplify the problem: $$u = 81 - x^2$$ Now, we need to find the derivative of \(u\) with respect to \(x\) to make a further substitution: $$\frac{d u}{d x} = -2x$$ Next, we need to rewrite the numerator in terms of \(u\), and eliminate the factor \(dx\). Notice that if we solve the expression for \(x^2\) and find its derivative $$x^2 = 81-u \Rightarrow\frac{d\left( x^2 \right)}{d x}=-\frac{du}{dx}=2x$$ We can then rewrite our integral in terms of \(u\).
02

Rewrite the integral in terms of \(u\)

We have: $$\frac{-1}{2}d u = x d x$$ Substituting, we get the following integral: $$-\frac{1}{2} \int \frac{u}{u^2} d u$$ Now, combine the \(u\) terms in the denominator: $$-\frac{1}{2} \int \frac{1}{u} d u$$
03

Integrate with respect to \(u\)

Now, we can easily integrate this expression in terms of \(u\): $$-\frac{1}{2} \int \frac{1}{u} d u = -\frac{1}{2} \ln|u| + C$$
04

Substitute back with the original variable \(x\)

Finally, we need to substitute back the original variable \(x\). Remember that \(u = 81 - x^2\). So our final answer is: $$-\frac{1}{2} \ln|81 - x^2| + C$$

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Most popular questions from this chapter

Many methods needed Show that \(\int_{0}^{\infty} \frac{\sqrt{x} \ln x}{(1+x)^{2}} d x=\pi\) in the following steps. a. Integrate by parts with \(u=\sqrt{x} \ln x.\) b. Change variables by letting \(y=1 / x.\) c. Show that \(\int_{0}^{1} \frac{\ln x}{\sqrt{x}(1+x)} d x=-\int_{1}^{\infty} \frac{\ln x}{\sqrt{x}(1+x)} d x\) and conclude that \(\int_{0}^{\infty} \frac{\ln x}{\sqrt{x}(1+x)} d x=0.\) d. Evaluate the remaining integral using the change of variables \(z=\sqrt{x}\) (Source: Mathematics Magazine 59, No. 1 (February 1986): 49).

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