Problem

Evaluate the integrals. $$\int \frac{1}{\sqrt{x^{…

01:23

Problem 27 Hard Difficulty

Evaluate the integrals.
$$\int \sqrt{x^{2}+16} d x$$

Answer

$\frac{1}{2} x \sqrt{16+x^{2}}$
$+8 \ln \left|\frac{1}{4} \sqrt{16+x^{2}}+\frac{x}{4}\right|+c$

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Calculus 1 / AB

Calculus 2 / BC

Calculus: Early Transcendental Functions

Chapter 6

Integration Techniques

Section 3

Trigonometric Techniques of Integration

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Video Transcript

Okay, so it's probably ready. 16 has for three Part two. And now we can x be equal to two. Or actually, that's four tangents of data. So we know that he x is it to actually for taking squared of data data, sir? Veering this will get the integral of, um, we can factor out square root of 16. That's four. And then we get the square root of tension squared. Plus one, which is tangents. We're better are actually that seeking squared. And then we have the X is four c gets squared of data, be data. Okay, so we got four cents. Lord a 16 in a row of tickets. Huge of data. You get out. Okay. Now we can use arm reduction for you. Love for sequins. That's equal, Teoh. 16 times he can't strip are three minus two. So that's seeking better. And then we have tangents of Biddle every minus one, which is to close 1/3. Actually, that's not being minus Tuesday. Um, three minus two is 1/3, minus one, which is to in a girl of chickens from theater. You dio okay. And we also know the ropes. Eakins So it's this right that out. That's 1/2 the natural log of chickens of data plus can engines of their own. And then we have plus E OK, And now let's find sequence intentions. So we know that if you draw out, try and go, that's X over four is equal to tensions if they don't. So our triangle has opposite over adjacent, and this high Pontin used is X squared plus four squared. Okay, so we know that Diggins data should be equal to in terms of for over the square root of X squared plus 16 and changes of data that's equal to Axl before.

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