Solving $\int_7^9 \frac{2}{9 + 16x^2}\,dx$ without using trigonometric substitution? How to evaluate $\int_7^9 \frac{2}{9 + 16x^2}\,dx$ without using trigonometric substitution?
I know how to do it with trigonometric substitution, but the problem I'm doing requires me to do it with algebra only. 
 A: The antiderivative involves an inverse trigonometric function, so whether a particular method counts as trigonometric is perhaps arguable:
Here's one option:
Partial fractions over $\mathbb{C}$
We can factor the denominator of the integrand over $\mathbb{C}$ as
$$9 + 16 x^2 = (3 i + 4 x)(-3 i + 4 x),$$
so we can decompose the integrand via partial fractions into a an expression
$$\frac{A}{3 i + 4 x} + \frac{B}{-3 i + 4 x}.$$
One can then integrate each term separately, then recombine the terms and apply identities to recover the antiderivative. (The involved identities, of course, probably require some trigonometry to prove.)
A: Note that
$$\frac2{9+16x^2}=\frac29\cdot\frac1{1+\left(\frac{4x}3\right)^2}\;,$$
and let $u=\frac43 x$. You should recognize the resulting integrand as a familiar derivative. (I wrote this before seeing your comment about the unavailability of $u$-substitution. I'm not sure that you can avoid it completely, at least without using less elementary techniques. Since this one is just a scalar multiple of $x$, perhaps it can be explained without too much difficulty.)
A: Here's another option:
Hyperbolic trigonometric substitution
This is perhaps cheating (especially from the complex point of view), but one can substitute
$$4 x = 3 \sinh t, \qquad dx = \frac{3}{4} \cosh t.$$ This transforms the (indefinite) integral to
$$\frac{1}{6} \int \text{sech } t \,dt,$$
and this integral appears on any good integral table.
A: Hint:
$$\int_7^9 \frac{2}{9+16x^2}dx=\frac{2}{9}\int_7^9 \frac{1}{1+(\frac{4}{3}x)^2}dx$$
And now let $\ \frac{4}{3}x=t$
A: Trigonometric substitution does not work(logically), unless there is a finite product of derivatives (I call this a virtual substitution, others call it u-substitution, but it is not a real substitution). The substitution theorem in integral calculus only works if the function that is being substituted is invertible and differentiable. Trig functions have infinitely many inverse functions in an arbitrary interval, which then violates the theorem. Unless you know that the derivative of arctan(u) is $\frac{u'}{1+u^2}$, which then gives the relation for the anti-derivative -- I cannot think of any elementary and purely algebraic manipulations in which you can prove this integral.
