Evaluate the integrals in Exercises 53–76.
57. ∫dx/(x√(2...

Question

Evaluate the integrals in Exercises 53–76.

57. ∫dx/(x√(25x²-2))

Accepted Answer

Welcome back everyone. Compute the indefinite integral dx divided by x, square root of 16 x 2 minus 3. We're going to solve this problem using U substitution. Let's suppose that U is equal to the square root of 16 x 2 minus 3. Let's remember that our next step is to find the derivative of u with respect to x, and in this case that's the derivative of a composite function square root of 16 x 2 minus 3, so we have to apply the chain rule. First of all, we're going to find the derivative of square root of 16 x 2 minus 3 with respect to 16 x 2 minus 3, which is 1 divided by 2 square root of that inner function 16 x 2 minus 3, and according to the chain rule we're multiplying by the derivative of 16 x 2 minus 3. Which is going to be 302 X. Divided by 2 square root of 16 x 2 minus 3, and we can also simplify, right? 32 divided by 2 gives us 16, so that's 16 x. Divided by square root of 16 x 2 minus 3 And from here, let's go ahead and solve for DX. What we can do is use cross multiplication, so that 16 x multiplied by dx is going to be Square root of 16 x 2 minus 3 DU And from here we can essentially. Sulfur DX by dividing both sides by 16 x so that DX is equal to square root of 16 x 2 minus 3. The EU Divided by 16 x. Additionally, notice that in the numerator we have u multiplied by d u, and for now we're going to leave x in terms of x. Well then, and now let's go ahead and substitute into the integral. We're going to have the integral of DX, which is U multiplied by DU divided by 16 x and we're also dividing by another x and square root of 16 x squared minus 3 which is Our you term, right? So now what we're going to do is simplify. Notice that we can cross out you and you. And we are basically integrating 1 divided by 16, we can factor out the constant, right? Integral of d u divided by x multiplied by x gives us x 2. So our next step is to rewrite x in terms of u. Let's go ahead and square both sides to get u2 equals 16 x2 minus 3. And basically what we're going to do is add 3 to both sides and divide it by 16, so that x 2 is equal to u2. Plus 3 divided by. 16. Therefore we get 1 divided by 16. Integral of d u divided by U 2 + 3 and then divided by 16. Notice that we have that double division. So essentially what we can do is write 16. In the numerator, and this allows us to cross out the constants. So basically this integral is equal to the integral of d u divided by u2 + 3. Notice that we can rewrite 3s square root of 3 squared and apply the formula for this integral, right? Remember the formula. According to the formula, we're going to get 1 divided by A. In this case, A is square root of 3 because we have an integral in the form of integral DU divided by U2 plus a 2. So we take 1 divided by the square root of 3. In verse of tangent of You divided by again a, so that'd be you divided by the square root of 3 plus a constant of integration c. Now, what is our next step? Well, essentially we are going to replace u with the actual expression in terms of x. So we're going to get 1 divided by a square root of 3. Inverse of tangent of Square root of 16 x square minus 3 divided by 3, right, we can combine the radical terms into a single expression plus a constant of integration c. And that's our final answer for this problem. Thank you for watching.

Evaluate the integrals in Exercises 53–76.57. ∫dx/(x√(25x²-2))

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Evaluate the integrals in Exercises 53–76.
57. ∫dx/(x√(25x²-2))