Find the indefinite integral.∫1x4x2−16dx\int_{}^{}\frac{1}{x\sqrt{4x^2-16}}dx

1 4 s e c − 1 ∣ x 2 ∣ + C \frac14sec^{-1}\left|\frac{x}{2}\right|+C

Find the indefinite integral. ∫ 1 x 4 x 2 − 16 d x \int_{}^{}\frac{1}{x\sqrt{4x^2-16}}dx

we're asked to find the indefinite integral of one over x times the square root of four x squared minus 16 d x. Now looking at this, seeing this x on the outside of my radical here, that's usually a good sign that we're going to be using this formula here that results in an inverse secant function. Now I'm going to go ahead and rewrite this integral slightly just so it's really clear what we're identifying u and a to be. So rewriting this as the integral of one over x times the square root of two x squared rather than that four x minus four squared, just rewriting those values. Now looking at this, it's clear that this would be my a value, that constant. But when I look to identify u, these two things should be the same based on my formula here, and they're not. That one on the outside of my radical is off by a factor of two, but I can fix that by just multiplying it by two. But if I'm multiplying by two in the denominator, I also need to multiply by two in the numerator so that I'm not actually changing the value of my integral. So now that I've done that here, I can go ahead and identify my U as being two X. Now, if U is equal to two X, what does that make DU? It makes it equal to two DX, which is exactly what I have right here. So with UA and DU identified, this is of the form one over the absolute value of U, even though we're missing those absolute value signs, that is okay here, times the square root of U squared minus A squared DU. Now this is going to be equal to one over a times the inverse secant of the absolute value of U over a. So going ahead and setting that up here, one over a, so that's one over four times the inverse secant of the absolute value of U over a. So that's two X over four and then plus C. Now we can do some simplification here. This two X over four, I can rewrite as just X over two. So making this one fourth inverse secant of the absolute value of x over two plus c. And this gives me my final answer here. Let us know if you have questions and I'll see you in the next one.

Find the indefinite integral. ∫ 1 x 4 x 2 − 16 d x \int_{}^{}\frac{1}{x\sqrt{4x^2-16}}dx

1 4 s e c − 1 ∣ x 2 ∣ + C \frac14sec^{-1}\left|\frac{x}{2}\right|+C

1 4 c s c − 1 ∣ x 2 ∣ + C \frac14csc^{-1}\left|\frac{x}{2}\right|+C

s e c − 1 ∣ x 2 ∣ + C sec^{-1}\left|\frac{x}{2}\right|+C

1 4 s e c − 1 ∣ x 4 ∣ + C \frac14sec^{-1}\left|\frac{x}{4}\right|+C

11. Integrals of Inverse, Exponential, & Logarithmic Functions

Integrals Involving Inverse Trigonometric Functions

Calculus

11. Integrals of Inverse, Exponential, & Logarithmic Functions

Integrals Involving Inverse Trigonometric Functions

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Multiple Choice

Find the indefinite integral.
$\int_{}^{} \frac{1}{x \sqrt{4 x^{2} - 16}} d x$

$\frac{1}{4} c s c^{- 1} ∣ \frac{x}{2} ∣ + C$

$s e c^{- 1} ∣ \frac{x}{2} ∣ + C$

$\frac{1}{4} s e c^{- 1} ∣ \frac{x}{2} ∣ + C$

$\frac{1}{4} s e c^{- 1} ∣ \frac{x}{4} ∣ + C$

Verified step by step guidance

Step 1: Recognize that the integral involves a square root of a quadratic expression in the denominator, specifically $ \sqrt{4x^2 - 16} $. This suggests a trigonometric substitution might be appropriate. Rewrite $ 4x^2 - 16 $ as $ 4(x^2 - 4) $ to simplify the expression.

Step 2: Use the substitution $ x = 2\sec(\theta) $, which implies $ x^2 - 4 = 4\sec^2(\theta) - 4 = 4(\sec^2(\theta) - 1) = 4\tan^2(\theta) $. This substitution simplifies the square root $ \sqrt{4x^2 - 16} $ to $ 2\tan(\theta) $.

Step 3: Compute $ dx $ using the substitution $ x = 2\sec(\theta) $. Differentiating gives $ dx = 2\sec(\theta)\tan(\theta) d\theta $. Substitute $ x $, $ dx $, and $ \sqrt{4x^2 - 16} $ into the integral.

Step 4: After substitution, the integral becomes $ \int \frac{1}{x\sqrt{4x^2 - 16}} dx = \int \frac{1}{2\sec(\theta) \cdot 2\tan(\theta)} \cdot 2\sec(\theta)\tan(\theta) d\theta $. Simplify the expression to $ \int \frac{1}{4} d\theta $.

Step 5: Integrate $ \int \frac{1}{4} d\theta $ to get $ \frac{1}{4}\theta + C $. Finally, use the substitution $ \theta = \sec^{-1}\left|\frac{x}{2}\right| $ to express the result in terms of $ x $. The final answer is $ \frac{1}{4}\sec^{-1}\left|\frac{x}{2}\right| + C $.

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Find the indefinite integral.∫1x4x2−16dx\(\int\)_{}^{}\(\frac{1}{x\sqrt{4x^2-16}\)}dx

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Find the indefinite integral.
$\int_{}^{} \frac{1}{x \sqrt{4 x^{2} - 16}} d x$