How do I solve this integral equation?

Question

Int: (x^3)*sqrt[(x^2)+3]

The final answer should be (1/5)[((x^2)+3)^(5/2)] - [(x^2)+3]^(3/2) + C

I really don't know how to go about solving this.

Answer 1

Integral [ x^3 sqrt(x^2 + 3) dx ]

Yeah; these aren't immediately intuitive, but hopefully after I answer this, it will be and you'll be on the lookout for these types of questions.

First, decompose x^3 into x^2 and x. What I'm going to do is move the x to the tail end of the integral, alongside dx.

Integral [x^2 sqrt(x^2 + 3) x dx ]

At this point, we use u-substitution.

Let u = x^2.
Then du = 2x dx, and
(1/2) du = x dx. {Note that x dx is the tail end of our integral, and we substitute accordingly.}

Integral ( u sqrt(u + 3) (1/2) du )

Now, pull the constant (1/2) out of the integral.

(1/2) Integral ( u sqrt(u + 3) du )

Note that sqrt(u + 3) is the same as (u + 3)^(1/2), so we have

(1/2) Integral ( u (u + 3)^(1/2) du)

To solve this, we use a second substitution.
Let v = u + 3. Then
v - 3 = u, so
dv = du

(1/2) Integral ( (v - 3) (v)^(1/2) dv )

Now, distribute the v^(1/2) over the (v - 3).

(1/2) Integral ( [v^(3/2) - 3v^(1/2)] dv )

Integrate as normal, using the reverse power rule.

(1/2) [ (2/5)v^(5/2) - 3(2/3)v^(3/2) ] + C

Distribute and simplify.

(1/5)v^(5/2) - v^(3/2) + C

Substitute v = u + 3.

(1/5)[u + 3]^(5/2) - [u + 3]^(3/2) + C

Now, substitute u = x^2.

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

Answer 2

let y = x^2 +3

dy = 2x dx

also x^2 = y - 3


using this in integral

Integral = int x^3 sqrt (x^2 + 3) dx

= int x^2 x sqrt (x^2 + 3) dx

= int x^2 sqrt (x^2 + 3) (x dx )

= int (y-3) sqrt (y) (1/2 dy)

= 1/2 {int y sqrt(y) dy- 3 int sqrt (y) dy }

= 1/2( int y^ 3/2 dy - 3 int y^ 1/2 }

= 1/2 { (y^5/2 / 5/2) - (3 y^3/2 / 3/2 ) } + c

= 1/5y^5/2 - y^3/2 + c

(sustituting for y)

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

Answer 3

well it is really simple u have to do the followings:
first let x^2+3=u so 2xdx=du so xdx=du/2
and x^2=u-3
now put it in ur integral so:
=> int[(u-3)sqrt(u)du/2]=
int[1/2(u^(3/2)-3u^(1/2)]
{and we now that
int(u^(n))=1/(n+1)*u^(n+1)}
so :
=1/2*[2/5 u^(5/2)-3*2/3u^3/2] =
by putting u=x^2+3 and simplifying we will have:
1/5*[(x^2+3)^1/5]-(x^2+3)^3/2
+c
which is ur answer simple and sweet

Answer 4

differentiate the two factors of the equation with understand to t (seem up Leibniz's imperative rule for the thank you to try this appropriate), then you have an subject-loose differential equation to sparkling up (use an integrating element, this feels like it is going to paintings out very cleanly given the presence of the two e^t and e^-t words). acquire the preliminary situation from the imperative equation itself, i.e. permit t = 0, then the above is f(0) = e^0 = a million. Make experience?