U substitution with integral

**Rstewart** · December 14th 2011, 01:05 PM

Integrate 4 + 2x(1+x^2)^3 dx from x=0 to 1

I tried u substitution with u = 1 + x^2 and du = 2x dx and got 7.75, but the answer in the back of the book says 19. What did I do wrong?

**Siron** · December 14th 2011, 01:13 PM

Your answer is correct.
http://www.wolframalpha.com/input/?i...dx+from+0+to+1

**Siron** · December 14th 2011, 01:15 PM

Your answer is correct.
integrate 4+2x(1+x^2)^3dx from 0 to 1 - Wolfram|Alpha

**Prove It** · December 14th 2011, 03:20 PM

Originally Posted by Rstewart

Integrate 4 + 2x(1+x^2)^3 dx from x=0 to 1

I tried u substitution with u = 1 + x^2 and du = 2x dx and got 7.75, but the answer in the back of the book says 19. What did I do wrong?

$\displaystyle \begin{align*} \int_0^1{4 + 2x\left(1 + x^2\right)^3\,dx} &= \int_0^1{4\,dx} + \int_0^1{2x\left(1 + x^2\right)^3\,dx} \\ &= \left[4x\right]_0^1 + \int_1^2{u^3\,du}\textrm{ after making the substitution }u = 1 + x^2 \implies du = 2x \\ &= 4\cdot 1 - 4\cdot 0 + \left[\frac{u^4}{4}\right]_1^2 \\ &= 4 + \frac{2^4}{4} - \frac{1^4}{4} \\ &= 4 + 4 - \frac{1}{4} \\ &= 7.75 \end{align*}$

Your answer is correct.

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