We can see that this is similar to a Pythagorean identity. So we're going to do trig substitution. We're going to make our triangle. The opposite side is going to be X ^2. The adjacent side is going to be 5. So our hypotenuse is going to be the square root of the X ^2 ^2 + 5 ^2, or what's actually down here in the denominator. So tangent Theta is going to be opposite over adjacent X ^2 / 5. Multiply each side by the five and we get 5 tangent Theta equaling X ^2. Taking the derivative of each side, we get 5 secant squared Theta D Theta equaling 2X DX, or five halves secant squared Theta D Theta equaling X DX. So if we come back here, there's our X DX. It's going to come out. We're going to replace it with five halves, secant squared Theta, D Theta in the denominator. We're going to get sqrt 25 plus instead of X to the 4th. We're going to take this X ^2 and square it. If I square one side, we're going to square the other. So I'm going to replace it with 25 tangent squared Theta. Factoring out a sqrt 25, that leaves US1 plus tangent squared Theta. Well, the square root of a secant squared Theta is going to be secant Theta and the fives are going to cancel. So this is really going to simplify down to 1/2 secant Theta D Theta. And we don't know how to evaluate this the way it is. So we're going to multiply by secant Theta plus tangent Theta over secant Theta plus tangent Theta. And our thought process there should be that. We do know how to evaluate the derivative of secant Theta tangent Theta. So if we multiply this out, we get secant squared Theta plus secant Theta tangent Theta D Theta all over secant Theta plus tangent Theta with the 1/2 in front. If we do AU sub, we let this denominator equal our U here. Then when we take the derivative we get DU equaling secant Theta tangent Theta plus secant squared Theta D Theta, which is that whole numerator. So this is really going to turn into 1/2 the integral DU over U, which is just one half the natural log of the absolute value of U + C back here. Remember U was tangent Theta plus secant Theta or secant Theta plus tangent Theta. So we have 1/2 the natural log, the absolute value of secant Theta plus tangent Theta plus C When we look at our right triangle up here, we're going to have 1/2 the natural log. Secant is hypotenuse over adjacent plus tangent is opposite over adjacent. So it simplifies into 1/2 the natural log square root X to the 4th plus 25 / 5 + X ^2 / 5 + C. Now we could do some interesting things here. We could realize that that really has a common denominator. So we could think about rewriting this as 1/2 natural logs, square root X to the 4th plus 25 plus X ^2 all over 5 + C and then that natural log of five is really just a constant. So we could rewrite this as 1/2 the natural log, the absolute value of X to the 4th plus 25 + X ^2 and it's in the denominator. So we could subtract 1 half natural log of 5 + C and we could combine this -1 half natural log of 5 + C and make it a new constant, maybe called C1. So a different answer would have been 1/2 the natural log of the absolute value of square root X to the 4th plus 25 + X ^2 minus some constant or plus some constant even. Because constants are just constants, they could be positive or negative. So two different answers. One, I would have been perfectly fine with leaving it here. If you realized that the numerators could be combined over a denominator, and the denominator was just a constant, you could actually simplify it a little more.