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JH

# Make a substitution to express the integrand as a rational function and then evaluate the integral.$\displaystyle \int \frac{x^3}{\sqrt{x^2 + 1}}\ dx$

## $$\frac{3}{10}\left(x^{2}+1\right)^{5 / 3}-\frac{3}{4}\left(x^{2}+1\right)^{2 / 3}+C$$

#### Topics

Integration Techniques

### Discussion

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##### Catherine R.

Missouri State University

##### Kristen K.

University of Michigan - Ann Arbor

##### Michael J.

Idaho State University

Lectures

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### Video Transcript

let's use the substitution to rewrite this integral as a rational function. So here, let's go ahead and take you to be this term in the denominator so you would be the cube root, X squared, plus one. Then we could go out in different share right now. But let's make this easier by Cuban, both sides give her that radical, then go ahead and take a derivative on in charge Differential. So we see that there's a DX and the original problems was going and solve this new expression for DX. So we'LL divide both sides by two ex and then we can say what exes By using this equation here. So ex player, you Q minus one, then here. If we want to use X, let's just take the square root So we have three use weird, you cued minus one one half power. We also have a two down there and then do you? So that's what we'LL use to replace the DX. Now we also have to deal with ex cubes and the denominator. So let's look at X cubed. We can get that by just Cuban full sides of this equation. So then we have you cubed minus one Then raising this to the three we have three over two is our power. So the integral X cube is you cubed minus one three Ask And then for the denominator, we see that we just have this term here and by Are you subs This's just you. So we could just put you down here And then we found the ex to be this expression over here on the right three u squared two You cubed minus one to the one half hour and then do you Now we'LL just go ahead and simplify the way that it's rich It's not. It's technically not a rational function until we cancel because of these one half hours. These are not polynomial sze. So let's go ahead and cross off So we have a three house Let's attract that one half hour We'Ll just get exponents of one Now we'LL have a rational function and further let's take off one of those use so we just have integral pull up the three half and then left over We have a U and then you cubed minus one and then the nominator is no longer there because it cancelled. So this we can go ahead and simplify. Yeah, you know, the fourth minus you and then was quiet and just use the power rules. Place three half you know, the five over five minus three halfs. Useful it over too. And now we are a constant of immigration, See? And the last thing to do here is to just replace you with X. So this is where we use the original definition of you. So we have three over ten, and now you to the fifth will become X squared, plus one. So originally, it was to the one third. But we're raising that to the fifth. So we get five thirds. And then here, three over four x squared, plus one. This time a raised were squaring it. So it's to the two thirds. Then we add our constant of integration. See? And that's your final answer.

JH

#### Topics

Integration Techniques

##### Catherine R.

Missouri State University

##### Kristen K.

University of Michigan - Ann Arbor

##### Michael J.

Idaho State University

Lectures

Join Bootcamp