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# Mathematics Integration?

Need some help with this question

Integrate 2x(4x+7)^8 dx

I'm getting lost in some of the working for this question

### 3 Answers

- husoskiLv 71 month agoFavorite Answer
The usual first place to look for a substitution is in the "messy part" of the integrand. Here, that's (4x + 7)^8, which cries out for a u=4x + 7 substitution; with x = (u-7)/4 and dx = du/4. You get:

∫ 2x (4x + 7)^8 dx = ∫ 2(u - 7)/4 * u^8 * du/4

= (1/8) ∫ (u - 7) u^8 du

= (1/8) ∫ (u^9 - 7u^8) du

= (1/8) [ u^10 / 10 - 7u^9 / 9] + C

= (u^9 / 8) [u/10 - 7/9] + C

= [(4x + 7)^9 / 8] * (1/90) [9(4x + 7) - 10(7)] + C

= [(4x + 7)^9 / 720] * (36x + 63 - 70) + C

= (4x + 7)^9 (36x - 7) / 720 + C

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- PhilipLv 61 month ago
J = Int'l[2x(4x+7)^8]dx. Put u = 4x+7. Then x =(u-7)/4 & dx = (1/4)du and

J = Int'l[f(u)]du, where f(u) = (1/2)(u-7)(1/4)u^8 = (1/8)(u-7)u^8.

Then 8J = int'l[u^9 -7u^8]du = (1/10)u^10 - (7/9)u^9 = u^9[u/10 -(7/9)] =

[(9u-70)/90]u^9 and J = [(9u-70)/720]u^9. 9u-70 = 9(4x+7) -70 = 36x -7.

Then J = [(36x-7)/720](4x+7)^9 + k, where k is an arbitrary constant of

integration.

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- DylanLv 61 month ago
This is an amazing site for this.

- James1 month agoReport
Thanks for that, will definitely be using this to check my answers in the future

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