# does $\int_{-1}^{1} x^3(x^2+1) dx = \frac{1}{2} \int_{-1}^{1} x^2(x^2+1) dx^2 = \frac{1}{2} \int_{-1}^{1} x(x+1) dx$?

I got this:

$$0 = \int_{-1}^{1} x^3(x^2+1) dx = \frac{1}{2} \int_{-1}^{1} x^2(x^2+1) dx^2 = \frac{1}{2} \int_{-1}^{1} x(x+1) dx \neq 0$$

I think the third equation is wrong, but what is the correct answer?

• Welcome to MSE. A question should be written in such a way that it can be understood even by someone who did not read the title. Dec 6, 2021 at 9:59

Your mistake is the last "$$=$$" sign: $$\frac{1}{2} \int_{-1}^{1} x^2(x^2+1) dx^2 = \frac{1}{2} \int_{-1}^{1} x(x+1) dx \neq 0$$ In fact, we did a substitution $$t=x^2$$ here, explicitly.-- But there is a mistake, as the inverse function of $$t=x^2$$ is not a one-value function on the interval $$x\in[-1,1]$$.
The correct is as below: \begin{align} \int_{-1}^1{x^2}(x^2+1)\mathrm{d}x^2=&\int_0^1{x^2}(x^2+1)\mathrm{d}x^2+\int_{-1}^0{x^2}(x^2+1)\mathrm{d}x^2 \\ (\color{green}{\text{use}\;t=x^2})=&\int_0^1{t}(t+1)\mathrm{d}t+\int_1^0{t}(t+1)\mathrm{d}t \\ =&\int_0^1{t}(t+1)\mathrm{d}t-\int_0^1{t}(t+1)\mathrm{d}t \\ =&0 \end{align} which avoids that $$x=\begin{cases} -\sqrt{t}\,\,/; x<0\\ \sqrt{t}\,\,/; x\ge 0\\ \end{cases}$$.