# Simple Integration by Substitution

I am just exercising integration by substitution and would like to solve the following integral:

$$\int (x^2+1)^3 dx$$

I substitute $x^2$ with y and hence $\frac{dy}{dx} = 2x$ and $dx = dy*2x$. Thus I get:

$$\int (y+1)^3 dy*2x$$

how do I continue from this point on, i.e. how do I get rid of the 2x?

• Put 2x under the integral, it doesn't get a free ride. – Loki Clock Apr 26 '13 at 21:10

Here, the substitution isn't appropriate, because, as you see, there's no factor of $x$ in the original integrand. It would have been a better choice to substitute $y = (x^2 + 1)$ if the integral had been $$\int x(x^2 + 1)^3\,dx$$ because then we'd have $\;dy = 2x dx\implies dx = \frac 12 dy,\;$ giving us a very nice integral to work with: $$\;1/2 \int y^3 dy$$

But, alas! We don't have that integral to work with. And there's not a really handy substitution to use that will simplify our work.

Instead, for this integral, try expanding the binomial (easy to do in this case), and use the power rule to integrate each term:

$$\int (x^2+1)^3 dx \quad = \quad \int (x^6 + 3x^4 + 3x^2 + 1) \,dx \quad = \quad\dfrac{x^7}{7} + \frac{3x^5}{5} + x^3 + x + C$$

• See: binomial expansion. Makes it easier to expand binomial raised to 3+. – Justin Apr 26 '13 at 21:14
• thank you, I know the binomial expansion works fine in this example, was wondering though how to use substitution. – user66280 Apr 26 '13 at 21:18
• Substitution works great...when it works and simplifies our work. But in this case, it substitution really isn't appropriate. – Namaste Apr 26 '13 at 21:23
• Note also, even if you had the integral, using your substitution: $$\int \frac{(y + 1)^3}{\sqrt y} \,dy$$, you'd still have to expand the binomial to integrate, account for the divition, and integrate the same number of terms that you'd have if you just proceeded, from the start, with expanding the binomial, no substitution. Plus, you'd need to back substitute if you use substitution. Hence, no reason to substitute, makes more work, not less. – Namaste Apr 26 '13 at 21:36
• @amWhy: Nice work! +1 – Amzoti Apr 27 '13 at 0:39

$\frac{dy}{dx} = 2x$ implies $dx = \frac{dy}{2x}$.

To get rid of the $x$, recognize that $x^2 = y$ implies $x = \pm \sqrt{y}$.

• Note that this substitution probably won't make your integral any easier. – Emily Apr 26 '13 at 21:08

You should not perform substitution here because the derivative $2x$ was not there to begin with, making for a difficult process. I suggest actually expanding out $(x^2+1)^3$ into a polynomial, which then is very easy to integrate.