# Simple Integral

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

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.

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

The simple substitution is u2=1+x2.
Might confuse him/her.

u=1+x^2

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...

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:

$$c\int f(x)dx$$

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

x.sqrt(1+x^2)