5
$\begingroup$

The problem is: $$ \int { (x+1)({ x }^{ 2 } } +2x{ ) }^{ 5 }dx $$

The next step given by WolframAlpha is $$\int { (x+1)({ x }^{ 2 } } +2x{ ) }^{ 5 }dx\\ =\quad \frac { 1 }{ 2 } \int { { u }^{ 5 }du } $$

(While I realize I am doing somthing worng)

The steps I am taking are : $ \int { (x+1)({ x }^{ 2 } } +2x{ ) }^{ 5 }dx$.

Let $u={ x }^{ 2 }+2x, \, du = 2x+2\, dx$. Then I see I can rewrite $du$ as $du = 2(x+1)dx$ giving me $\frac { du }{ 2 } = (x+1)\, dx$.

I can now see where the $ \frac { 1 }{ 2 } $ is coming from, but I cannot seem to visualize the next steps to get to

$$ \int { (x+1)({ x }^{ 2 } } +2x{ ) }^{ 5 }dx=\frac { 1 }{ 2 } \int { { u }^{ 5 }du } $$

$\endgroup$
3
  • 1
    $\begingroup$ You are basically done. You just need to substitute $x^2+2x=u$ and $(x+1)dx=\frac12du$. $\endgroup$
    – sbares
    May 4, 2016 at 6:04
  • 1
    $\begingroup$ What happens when you replace $(x^2+2x)^5$ with $u^5$ and $(x+1) \mathrm{d}x$ with $\dfrac{1}{2} \mathrm{d}u$? $\endgroup$ May 4, 2016 at 6:05
  • $\begingroup$ Please don't put large pieces of text in LaTeX. Also, you use the curly braces unnecessarily many times. $\endgroup$
    – wythagoras
    May 4, 2016 at 6:12

2 Answers 2

2
$\begingroup$

You already state in your question that you found that $$\frac{\mathrm{d}u}{2} = (x+1)\mathrm{d}x$$

when doing the substitution $u=x^2+2x$. When doing this substitutiuon on the integral, we get $$\int (x+1)(x^2+2x)^5 \mathrm{d}x = \int (x+1)u^5 \mathrm{d}x = \int u^5 (x+1)\mathrm{d}x = \int u^5 \frac{\mathrm{d}u}{2} = \frac{1}{2} \int u^5 \mathrm{d}u$$

$\endgroup$
2
$\begingroup$

Since $(x^2 + 2 x)^5 = u^5$, we can use the equation $(x + 1) \,dx = \frac{1}{2} du$ to write the integral as $$\int \underbrace{(x^2 + 2 x)^5}_{u^5} \,\underbrace{(x + 1) \,dx}_{\frac{1}{2} du} = \int u^5 \cdot \frac{1}{2} du = \frac{1}{2} \int u^5 \,du .$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .