How do I integrate $\int\frac{x^3}{\sqrt{1+x^2}}$ by parts?

Question

So I've just recently begun integration and now we're doing integration by parts. We've been told about ILATE (some sort of acronym to helps us remember which function we integrate and which to differentiate).

But that aside can we perform by part integration for two functions of same type? For example,

$$\int\frac{x^3}{\sqrt{1+x^2}}.$$

In the above problem, Can I consider $x^3$ as one part and $\dfrac{1}{\sqrt{1+x^2}}$ as the other part? Sine then if I chose to integrate the second function, there is a standard integration and then differentiating $x^3$ is no big deal.

Am I thinking right? Or is there some fundamental mistake in my approach?

Answer 1

The trick to using integration by parts is to be able to resolve the resulting integral more easily than one you started with.

You view the original integrand as a product of two terms. Then you think about which term to differentiate and which to integrate. The point is that you'll need to integrate the new product of the differentiated and integrated terms easily. That's, in a sense, the "limiting step" in IBP.

With your proposed method you need to integrate something of the form $x^2 \arctan x$, which is hardly easier than what you started with.

Think slightly more creatively. $x^3 = x^2 \cdot x$. Now $\frac x{\sqrt {1+x^2}} $ is easy to integrate, think of the form $g'(x) \cdot f(g(x)) $. Obviously $x^2$ is trivial to differentiate. What's most important is that the product you end up with, of the form (ignoring constant multipliers) $x\cdot \sqrt {1+x^2}$ is also easy to integrate - again, think of $g'(x) \cdot f(g(x)) $.

That's the key. Break up the integrand into a product of two terms. Get creative in how you do this. You have to differentiate one term, integrate the other and the product needs to be easily integrable.

Answer 2

Put $1+x^2=t^2$ and $2xdx=2tdt\Longrightarrow xdx=tdt$

So integral $I =\int \frac{x^2}{\sqrt{1+x^2}}\cdot xdx=\int\frac{t^2-1}{t}\cdot tdt=\int (t^2-1)dt$

Answer 3

$$\int\dfrac{x^3}{\sqrt{1+x^2}}\ dx$$

$$=\int\dfrac{x(1+x^2)-x}{\sqrt{1+x^2}}\ dx$$ $$=\int \left(x\sqrt{1+x^2}-\dfrac{x}{\sqrt{1+x^2}}\right)\ dx$$ $$=\frac12\int \sqrt{1+x^2}\ d(1+x^2)-\frac12\int\dfrac{d(1+x^2)}{\sqrt{1+x^2}}$$ $$=\frac13(1+x^2)^{3/2}-\sqrt{1+x^2}+C$$

Answer 4

Maybe, it's better to do another integration technique. If you do \

$$u=x^2+1\Rightarrow du=2xdx,$$

so,

$$\dfrac{1}{2}\displaystyle\int \dfrac{u-1}{\sqrt{u}}du$$

that's easy to solve.

Answer 5

In fact, the aim of choosing which function to differentiate and which one to integrate is to simplify as much as we can. When you look at the function $\frac{1}{\sqrt{1+x^2}}$ you can see it's a little similar to a derivative of another function, which is the case here $\sqrt{1+x^2}'=\frac{x}{\sqrt{1+x^2}}$. So now we can write: $\frac{x^3}{\sqrt{1+x^2}}=x^2\times\frac{x}{\sqrt{1+x^2}}$. Now we can see very easily the choice you need to make, you differentiate $x^2$ and you integrate $\frac{x}{\sqrt{1+x^2}}$