1. ## Integral.

Hello.
evaluate:

$\int \sqrt{x^2+16} dx$.
Without using Trig/Hyperbolic Substitutions.
And you must use the substitution method only.
I believed, its impossible to solve it.

2. The way I was taught to answer integrals:
$\sqrt{x^2+16} dx$

$
=(x^2+16)^\frac{1}{2}dx
$

$
u = x^2+16 =>
du = 2x
$

$
=\frac{1}{2}\int(x^2+16)^{\frac{1}{2}}[2xdx]
$

$
=\frac{1}{2}*
\frac{u^{3/2}}{\frac{3}{2}}
$

$
= \frac{1}{3}(x^2+16)^{3/2}
$

The way I was taught to answer integrals:
$\sqrt{x^2+16} dx$

$
=(x^2+16)^\frac{1}{2}dx
$

$
u = x^2+16 =>
du = 2x
$

$
=\frac{1}{2}\int(x^2+16)^{\frac{1}{2}}[2xdx]
$
Unfortunately, you cannot just put that "x" into the integrand!

$
=\frac{1}{2}*
\frac{u^{3/2}}{\frac{3}{2}}
$

$
= \frac{1}{3}(x^2+16)^{3/2}
$
No, the derivative of $\frac{1}{3}(x^2+ 16)^{3/2}$ is
$(3/2)(1/3)(x^2+ 16)^{1/2} (2x)= x(x^2+16)^{1/2}$, not just $(x^2+ 16)^{1/2}$

I doubt that you were taught that way!

Ted, I don't believe there is any way to do that other than either the trig or hyperbolic substitutions.

Of course, you could always look it up in a table of integrals. Does that count?
http://www.integral-table.com/

4. Originally Posted by Ted
Hello.
evaluate:

$\int \sqrt{x^2+16} dx$.
Without using Trig/Hyperbolic Substitutions.
put $x=4u$ and the integral becomes $16\int\sqrt{u^2+1}\,du.$

put $t=\sqrt{u^2+1}+u$ then $\frac1t=\sqrt{u^2+1}-u,$ thus $\frac12\left(t+\frac1t\right)=\sqrt{u^2+1}$ and $\frac12\left(t-\frac1t\right)=u$ so i think you can finish it.