How do i integrate the following function?

Question

x(x+1)^1/2

I use the substitution du = 1 dx,
but from that point i have no idea how to integrate the function.

Please provide an explanation thanks.

Answer 1

Integrate by parts.

That is, you know d/dx(ab) = da/dx(b) + a(db/dx), or that

da/dx(b) = d/dx(ab) - a(db/dx)

So . . . let a = int((x+1)^(1/2) = 2/3(x+1)^(3/2), and b = x

then the integral is:

int(da/dx(b) = d/dx(ab) - a(db/dx))

= ab - int(a(db/dx))

2/3x(x+1)^(3/2) - int(2/3(x+1)^3/2)

=2/3x(x+1)^(3/2) - 4/15(x+1)^5/2

We can take the derivative to check:

=x(x+1)^(1/2) + 2/3(x+1)^(3/2) - 2/3(x +1)^(3/2)

=x(x + 1)^(1/2)

There you go.

Answer 2

The problem with this integral is the "complicated" expression inside the square root. In order to simplify the expression inside the square root, let's try the substitution u = x + 1.

Then
du = dx and x = u - 1

Substituting this into the integral gives
integral( (u-1) u^1/2 du)

multiplying out the integrand gives
integral ( u^(3/2) - u^(1/2) du )

which is an easier integral to evaluate.

Good Luck!

Answer 3

x(x+1)^1/2
let u=x+1
du=dx
integral of((u-1)*u^1/2 du)
integral of (u^3/2-^1/2 du)
2/5 u^5/2 -2/3 u^3/2 +C
2/5 (x+1)^5/2 -2/3* (x+1)^3/2 +C

Answer 4

Int sqrt(sixteen-x²)dx = Int (sixteen-x²)/sqrt(sixteen-x²)dx= Int sixteen/sqrt(sixteen-x²) dx - Int x²/sqrt(sixteen - x²)dx = 16Int a million/sqrt(sixteen-x²)dx + a million/2Int x * [sqrt(sixteen-x²)]`dx = sixteen/8 ln|(x-4)/(x+4)| +a million/2xsqrt(sixteen-x²) - a million/2Int sqrt(sixteen-x²)dx Intsqrt(sixteen-x²)dx+a million/2Intsqrt(sixteen-x²)=2ln...? =3/2 Int sqrt(sixteen-x²)dx = 2ln|(x-4)/(x+4)| - a million/2xsqrt(sixteen-x²) +C so Int sqrt(sixteen-x²)dx = (4ln|(x-4)/(x+4)| - xsqrt(sixteen-x²) + 2C) / 3 ** formula used a million)Int a million/a²-x² dx = (a million/2a)*ln| (x-a)/(x+a) | on your case you had Int a million/sixteen-x² so a=4 2) [sqrt(sixteen-x²)]` = (sixteen-x²) / 2sqrt(sixteen-x²)= = - 2x/2sqrt(sixteen-x²) = -x/sqrt(sixteen-x²) 3.)integration via areas **