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Chapter 1: Problem 19

Evaluate the following integrals: $$ \int \frac{(x+1) \sqrt{x+2}}{\sqrt{x-2}} d x $$

Short Answer

Expert verified
Question: Evaluate the integral of the given function: $$\int \frac{(x+1) \sqrt{x+2}}{\sqrt{x-2}} d x$$ Answer: The integral of the given function is $$\int \frac{(x+1) \sqrt{x+2}}{\sqrt{x-2}} d x = 2\left(\frac{(\sqrt{x - 2})^7}{7} + \frac{7(\sqrt{x - 2})^5}{5} + 4(\sqrt{x - 2})^3\right) + C$$

Step by step solution

01

Substitution

Let's perform the substitution \(u = \sqrt{x - 2}\). This means that \(u^2 = x - 2\), so \(x = u^2 + 2\). Taking the derivative of \(x\) with respect to \(u\), we have \(dx = 2u\,du\). Now we will rewrite the given integral in terms of \(u\), and we can find the antiderivative with respect to \(u\). $$ \int \frac{(x+1) \sqrt{x+2}}{\sqrt{x-2}} d x = \int \frac{((u^2 + 2) + 1) \sqrt{(u^2 + 2) + 2}}{\sqrt{u^2}} 2u \, du $$
02

Simplify the Expression

Now we simplify the expression inside the integral: $$ \int \frac{(u^2 + 3)u(u^2 + 4)}{u} 2u \, du = \int 2(u^2 + 3)(u^4 + 4u^2) \, du $$ After multiplying the two brackets, we have: $$ \int 2(u^6 + 4u^4 + 3u^4 + 12u^2) \, du $$
03

Perform the Integration

Now, we can perform the integration by finding the antiderivative of the expression inside the integral: $$ \int 2(u^6 + 7u^4 + 12u^2) \, du = 2\left(\frac{u^7}{7} + \frac{7u^5}{5} + 4u^3\right) + C $$
04

Unsubstitute and Evaluate

Now we will resubstitute the \(x\) back into our antiderivative for \(u\) by using the relationship we found in Step 1: $$ 2\left(\frac{(\sqrt{x - 2})^7}{7} + \frac{7(\sqrt{x - 2})^5}{5} + 4(\sqrt{x - 2})^3\right) + C $$ We have now found the antiderivative of the given integral: $$ \int \frac{(x+1) \sqrt{x+2}}{\sqrt{x-2}} d x = 2\left(\frac{(\sqrt{x - 2})^7}{7} + \frac{7(\sqrt{x - 2})^5}{5} + 4(\sqrt{x - 2})^3\right) + C $$

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Most popular questions from this chapter

Assuming that \(\int\left(\mathrm{e}^{\mathrm{x}} / \mathrm{x}\right) \mathrm{d} \mathrm{x}\) is not elementary (a theorem of Liouville), prove that \(\int 1 / \ln \mathrm{x} \mathrm{dx}\) is not elementary.

Evaluate the following integrals: (i) \(\int \frac{x d x}{x^{4}-x^{2}-2}\) (ii) \(\int \frac{d x}{x\left(a+b x^{2}\right)^{2}}\) (iii) \(\int \frac{\mathrm{x}}{\left(\mathrm{x}^{2}+2\right)\left(\mathrm{x}^{2}+1\right)} \mathrm{dx}\) (iv) \(\int \frac{\left(1-x^{2}\right) d x}{x\left(1+x^{2}+x^{4}\right)}\)

Evaluate the following integrals: $$ \int \frac{x^{2}+2 x-1}{2 x^{2}+3 x+1} d x $$

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