# Where am I going wrong in this integration

I am trying to do this integration but somehow I don't get the right answer:

$$\int(4(2x+1)^7)dx$$ $$u=2x+1 \space \therefore \dfrac{du}{dx}=2 \space \therefore dx=\dfrac{1}{2}du$$ $$4\times 2\int u^7\times \dfrac{1}{2}du$$ $$4\int u^7du = 4\dfrac{u^8}{8}=\dfrac{1}{2}(2x+1)^8+C$$ However it should be $\dfrac{1}{4}(2x+1)^8 + C$

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In line 3 you have a factor of 2 that doesn't belong there. Line 3 should be $4$ times the integral, not $4$ times $2$ times the integral. –  Dave L. Renfro Feb 8 '13 at 18:59
The $2$ product you pulled out is not correct. Look at that step again, you have a 1/2 term there. Regards –  Amzoti Feb 8 '13 at 19:00
Shouldn't it be there to cancel the 1/2 i put in, in the integral? –  user61518 Feb 8 '13 at 19:01
From line 2 to line 3 you're simply replacing various $x$-expressions (such as $2x + 1$ and $dx$) with their equivalent $u$-expressions. Multiplying and dividing by $2$ is part of a slightly different way of carrying out the same thing you're doing. –  Dave L. Renfro Feb 8 '13 at 19:05
Thanks I understood it now! –  user61518 Feb 8 '13 at 19:09

This is intended to clarify my first two comments.

Let's omit the factor of $4$, which is unrelated to your question. Then you want to evaluate

$$\int(2x+1)^7\;dx$$

Here's one way to view substitution. We know that

$$\int(boop)^7\;d(boop) \; = \; \left(\frac{1}{8}\right)(boop)^8 + C$$

A version of this that is close to what you have is

$$\int(2x+1)^7\;d(2x+1)$$

However, you have $dx$ and the integral I just wrote has $d(2x+1)$. But since $d(2x+1) = 2\,dx,$ we can get $d(2x+1)$ to show up this way:

$$\int(2x+1)^7\;dx \; = \; \int(2x+1)^7\;\left(\frac{1}{2}\right)\cdot 2\,dx$$

$$= \; \int(2x+1)^7\;\left(\frac{1}{2}\right)\cdot d(2x+1) \; = \; \frac{1}{2}\int(2x+1)^7 \; d(2x+1)$$

Now here's how your $u$-substitution method works in light of what I did above. You want to get

$$\int u^7 \,du$$

to show up. Obviously, $u = 2x+1$, so to get $du$ to appear we need to see what $du$ is. As you've shown, $du = 2\,dx.$ At this point you can either multiply and divide by $2$ to get $2\,dx$ to show up (the approach I took above), or you can simply substitute $\frac{1}{2}du$ in place of $dx$ (what you did). But if you choose the direct substitution method, you don't have to multiply by $2$ (where you made your error), since you're just replacing equals with equals.

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In case anyone was wondering, I used "boop" because "stuff" (what is usually used in situations like this when one wants to be eye-catching) because LaTeX renders "stuff" like this $stuff$ (extra space around each $f$), which I didn't like the appearance of. –  Dave L. Renfro Feb 8 '13 at 20:13
Boop d boop, reminds me of Betty Boop. –  Baby Dragon Feb 8 '13 at 20:15
@Baby Dragon: When I was thinking of a replacement for "stuff", the first thought (2 seconds or less) in my mind was "boop". I went with it, thinking I'd change it (when I got done with the rest of the post) if I didn't like it or had a better idea. However, it had some charm to it and nothing better immediately came to my mind, so "boop" it was. Also, none of the letters $u,$ $x,$ or $d$ appeared in it . . . –  Dave L. Renfro Feb 8 '13 at 20:18
+1 and haha: "...my mind was boop." –  draks ... Feb 27 '13 at 6:38

Let me solve this question step-by-step for you.

$$\int4(2x+1)^7dx$$

$$u=2x+1$$

$$\frac{du}{dx}=2 \therefore dx=\frac12du$$

$$\therefore \int4(2x+1)^7dx=4\int u^7 * \frac12du=2\int u^7 du$$

This becomes:

$$2(\frac{u^8}8) + c=\frac{u^8}4+c$$

$$=\frac{(2x+1)^8}4+c$$

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@Julian Kuelshammer thanks for the edit! –  Ayush Khemka Feb 9 '13 at 11:02

In general, for a derivable function $\,f(x)\,$ , we have

$$\int f'(x)f(x)^ndx=\frac{f(x)^{n+1}}{n+1}+C$$

In our case, $\,f(x)=(2x+1)\,\,,\,\,f'(x)=2\,$ , so:

$$\int 4(2x+1)^7\,dx=2\int(2\,dx)(2x+1)^7=2\frac{(2x+1)^8}{8}+C=\frac14(2x+1)^8+C$$

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Yup, a typo. Thanks for taking it down. –  DonAntonio Feb 9 '13 at 14:48