I can't figure out how to do this but Ive been assured its easily doable.
Calculate: (int) 2x(4x+7)^{8 }dx
So I'm assuming we try to get 4x on its own and find 2x in terms of u?
A solution to this would help me heaps here I just cant see it!
I can't figure out how to do this but Ive been assured its easily doable.
Calculate: (int) 2x(4x+7)^{8 }dx
So I'm assuming we try to get 4x on its own and find 2x in terms of u?
A solution to this would help me heaps here I just cant see it!
$\displaystyle \int 2x(4x+7)^8 \, dx$
$\displaystyle u = 4x+7$
$\displaystyle x = \frac{u-7}{4}$
$\displaystyle dx = \frac{du}{4}$
$\displaystyle \int 2x(4x+7)^8 \, dx$
$\displaystyle \int \frac{u-7}{8} \cdot u^8 \, du$
$\displaystyle \frac{1}{8} \int u^9 -7u^8 \, du$
can you finish?
note you don't have to resort to u-substitution, you COULD just multiply out (4x+7)^{8}, and then multiply every term by 2x, and then integrate term-by term. this is kind of "the hard way", u-substitutions are meant to make your life EASIER (short-cuts are good, right?).
linear substitutions u = ax+b are always "do-able", we just have a "fudge factor" of a constant to worry about:
du = a dx, so
dx = (1/a) du <--take the 1/a outside the integral sign.