How do I Integrate x(1+x)^(1/2)?

Question

It was originally written as evaluate the integral of x(squareroot(1+x) but its the same thing either way.

I was thinking integration by parts, but its not working. I don't want or need the answer I just want someone to put me on track and tell me which technique or techniques I should use. Thank you.

Answer 1

These integrals are a bit tricky; let me show you an example to demonstrate how things work with substitution.

Example #1: Integrate x(x-3)^100

If we were to expand (x-3)^100, we would be nuts. In theory, we could, and then integrate each term, but that would be unnecessarily time consuming. Here's what we do instead.

Integral (x (x-3)^100) dx

Let u = x - 3
u - 3 = x
Therefore, du = dx

So now our integral becomes

Integral ( (u - 3)u^100)du

Which we can now expand to

Integral ( u^(101) - 3u^100 ) du

And now our integral just became a whole heck of a lot easier.

[u^(102)]/102 - 3[ (u^(101))/101 ] + C

And then we just replace u again.

[(x-3)^(102)]/102 - (3/101)[ (x-3)^(101) ]

What we're going to do is the exact same thing for your question, which is

Integral (x (1+x)^(1/2))dx

Let u = 1+x
Then u - 1 = x, and
du = dx

Therefore, our integral becomes

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

Now, we can actually multiply it out, resulting in,

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

And we use the reverse power rule again for these, getting

(2/5)u^(5/2) - (2/3)u^(3/2) + C

Replacing u = 1+x,

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

You can do all sorts of funky stuff to reduce that, but I'll leave it up to you.

Answer 2

You can do this by substitution. Let u = x+1. Then x = u-1, du = dx. The integral becomes ∫(u-1)√u du, or ∫[u√u - √u]du = ∫√u^3/2 du- ∫u^1/2du. You can then use your standard integration of powers of a variable.

Answer 3

Integration by parts can be used, simply use u(x) = x and
v'(x)=(1+x)^(1/2) , i.e x is to be differentiated.

Else use the simple substitution u=1+x
then it becomes
int. (u-1)Vu du = int. u^(3/2) - u^(1/2) du

Answer 4

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Answer 5

please put 1+x =t^2;then dx= 2tdt
x= t^2-1

your equation becomes:
integral((t^2-1).t.2t)dt which is eariesr to solve.
finally you should get after substituing valve of t in the final answer
2{(1+x)^5/5 -(1+x)^3/3} +c

Answer 6

integration by parts works
req integral = x *integral[(1+x)^0.5] - integral[ 1* integral[(1+x)^0.5]
=[x(1+x)^1.5]/1.5 - integral[ (1+x)^1.5]/1.5
=[ x(1+x)^1.5 ]/1.5 - [1/(1.5*2.5)] * (1+x)^2.5

Answer 7

1 * (1 / 2) = 0.5