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I have to make the integration with the substitution rule:

$$\int {(x^2-1})^2(2x)dx$$

$$u=x^2-1$$

$$\int {\sqrt{u}}du=\frac{2u^{3/2}}{3}$$

$$\frac{2(x^2+1)^{3/2}}{3}(2x)$$

My question is:

Do I need to integrate de las part of the expression: the 2x? $$\int {\sqrt{u}}du\int {2x}dx=\frac{2u^{3/2}}{3}x^2$$

$$\frac{2(x^2+1)^{3/2}}{3}(x^2)$$

Haizum Skallah
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    As AWertheim implied, the substitution you were asked to make does not correspond to the original integral at all. It looks like a different problem altogether. Follow AWertheim's advice. – Adrian Keister Aug 07 '13 at 02:08

2 Answers2

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You have a slight error. We make the substitution

$$u = x^{2} -1; du = 2xdx$$

Then our integral becomes:

$$\int u^{2}du = \frac{1}{3}u^{3} + C = \frac{1}{3}(x^{2}-1)^{3} + C$$

I leave it to you to compute the final algebra steps.

Alex Wertheim
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if you dont want to do any substitution,in this question you can expand the square. $$\int (x^2-1)^2(2x)\;dx$$ $$\int (x^4+1-2x^2)(2x)\;dx$$ $$\int (2x^5+2x-4x^3)\;dx$$ now you can easily solve this integration.

iostream007
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  • (Thanks) I knew that, but I was asked to do it by "substitution method". Anyway when I have a product like integration of $$\int{ab}$$ where a need to be substituted and b can be treated normally whats the procedure for b ones you made the substitution for a? – Haizum Skallah Aug 07 '13 at 10:56