Integrating x√(1+x²) Without Trig Functions

[rwinston]

Hi

I am trying to integrate

x \sqrt{1+x^2}dx

by parts...but it seems to involve trigonometric functions - is it possible to solve this integral without using trig functions?


Thx

 

[d_leet]

Why are you trying to use integration by parts? This is a very simple integral if you use an appropriate substitution.

 

[mathman]

The simple substitution is u²=1+x².

 

[Feldoh]

Mathman's solution is the easiest, but you probably could do it by parts.

 

[rocomath]

The simple substitution is u²=1+x².

Might confuse him/her.

Just let u = radican.

u=1+x^2

 

[rwinston]

Hmm. Ok if u = 1 + x^2

du = 2x dx

\int x \sqrt{1+x^2}dx = \frac{1}{2}\int \sqrt{1+x^2}2xdx

=\frac{1}{2} \int\sqrt{u}du

=\frac{1}{3} (1+x^2)^{\frac{3}{2}}

But if I take u^2=1+x^2

u = \sqrt{1+x^2}

du = \frac{x}{\sqrt{1+x^2}}dx

\int{x\sqrt{1+x^2}dx} = \sqrt{1+x^2}\int{\sqrt{1+x^2}\frac{x}{\sqrt{1+x^2}}dx}

=\sqrt{1+x^2}\int{u du}

=\sqrt{1+x^2} \frac{{\sqrt{1+x^2}}^2}{2}

=\frac{(1+x^2)^{\frac{3}{2}}}{2}

I must be making a mistake somwhere...

 

[rocomath]

Hmm. Ok if u = 1 + x^2

du = 2x dx

\int x \sqrt{1+x^2}dx = \frac{1}{2}\int \sqrt{1+x^2}2xdx

=\frac{1}{2} \int\sqrt{u}du

=\frac{1}{3} (1+x^2)^{\frac{3}{2}}

But if I take u^2=1+x^2

u = \sqrt{1+x^2}

du = \frac{x}{\sqrt{1+x^2}}dx

\int{x\sqrt{1+x^2}dx} = \sqrt{1+x^2}\int{\sqrt{1+x^2}\frac{x}{\sqrt{1+x^2}}dx}

=\sqrt{1+x^2}\int{u du}

=\sqrt{1+x^2} \frac{{\sqrt{1+x^2}}^2}{2}

=\frac{(1+x^2)^{\frac{3}{2}}}{2}

I must be making a mistake somwhere...

No, he meant to use u^2

So ...

u^2=1+x^2 \rightarrow udu=xdx

And remember that you can only pull out constants:

c\int f(x)dx

 

[rootX]

wow, substitutions are so much mess. Try this way:

x.sqrt(1+x^2)

think about sqrt(1+x^2)
it has integral
(1+x^2)^3/2 --- [ref 1]

it has derivative 3x (1+x^2)^0.5

But, in original thing we only have x.sqrt(1+x^2)

so, you need 1/3 with [ref 1]