what is the integral of x(5x^2-4)^(1/2) dx?

Question

help me pleez

Answer 1

So you want to solve

Integral (x (5x^2 - 4)^(1/2)) dx

The first thing I'll do in order to make something obvious (this is a skippable step; only meant to teach you) is that I'm going to move the x next to the dx. Then, we will get

Integral ( (5x^2 - 4)^(1/2) x) dx

Here's where we use substitution.
Let u = 5x^2 - 4. Then
du = 10x dx, and
(1/10) du = x dx

Now, note that we have our "x dx" in our integral, and thus can completely replace it with (1/10) du. We also substitute our u = 5x^2 - 4 normally, and we get

Integral ( u^(1/2) (1/10))du

It's always a good idea to pull out constants before solving the integral. Whenever we have constants inside an integral, we can pull it outside of the integral (side note: this is true for limits and derivatives as well)

(1/10) * Integral (u^(1/2))du

Now, this is an easy integral to solve; it's just the reverse power rule for derivatives. This will give us

(1/10) * [(2/3) u^(3/2) ] + C

Now, we plug u = 5x^2 - 4 back, to get

(1/10) * [(2/3) [5x^2 - 4]^(3/2)] + C

We can simplify this slightly, by multiplying the fractions out.

(2/30) [5x^2 - 4]^(3/2) + C

And we can reduce.

(1/15) [5x^2 - 4]^(3/2) + C

Answer 2

Let u = 5x² - 4

du = 10xdx

==> dx = (1/10)xdx

==> ∫(5x² - 4)dx = (1/10)∫√u du
==> (1/10)(2/3)u^(3/2) + C
==> (1/15)(5x² - 4)^(3/2) + C

(done)

Answer 3

the answer is (1/15)(5x^2-4)^(3/2)+constant
Do your homework !

Answer 4

4/5
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