Question

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

Short answer

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

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 \(u\). $$ \int \frac{(x+1) \sqrt{x+2}}{\sqrt{x-2}} d x = \int \frac{((u^2 + 2) + 1) \sqrt{(u^2 + 2) + 2}}{\sqrt{u^2}} 2u \, du $$

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: $$ \int 2(u^6 + 4u^4 + 3u^4 + 12u^2) \, du $$

Perform the Integration

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

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: $$ \int \frac{(x+1) \sqrt{x+2}}{\sqrt{x-2}} d x = 2\left(\frac{(\sqrt{x - 2})^7}{7} + \frac{7(\sqrt{x - 2})^5}{5} + 4(\sqrt{x - 2})^3\right) + C $$