Evaluate the integral $\int x^{3} \sqrt{x^{2}+7} d x$
Answer
So, the integral \(\int x^{3} \sqrt{x^{2}+7} d x\) is equal to \(\boxed{\frac{2}{5}x^{4}\sqrt{x^{2} + 7} + \frac{14}{15}x^{2}\sqrt{x^{2} + 7} - \frac{196}{15}\sqrt{x^{2} + 7} + C}\).
Step 1 ：Let's start by setting \(u = x^2 + 7\).
Step 2 ：Then, we find that \(du = 2x dx\).
Step 3 ：However, we have \(x^3 dx\) in our integral, so we need to manipulate our substitution a bit.
Step 4 ：We can express \(x^2\) in terms of \(u\) and then substitute it back into the integral.
Step 5 ：Doing this, we find that the integral becomes \(\frac{2}{5}x^{4}\sqrt{x^{2} + 7} + \frac{14}{15}x^{2}\sqrt{x^{2} + 7} - \frac{196}{15}\sqrt{x^{2} + 7}\).
Step 6 ：Finally, we add the constant of integration \(C\) to our result.
Step 7 ：So, the integral \(\int x^{3} \sqrt{x^{2}+7} d x\) is equal to \(\boxed{\frac{2}{5}x^{4}\sqrt{x^{2} + 7} + \frac{14}{15}x^{2}\sqrt{x^{2} + 7} - \frac{196}{15}\sqrt{x^{2} + 7} + C}\).
