What is the integral of 1/(9x^2 +16)?

Question

How do you do it? I'm stuck. Thanks!

Answer 1

Integral of 1/(x^2+a^2) = 1/a* tan(inverse)x/a

Here we take 9 common in denominator and we get 'a^2' as 16/9 and a = 4/3

Thus ans is (1/9)*(3/4)*tan(inverse)3x/4

Answer 2

Integral (1 / (9x^2 + 16)) dx

You have to use trigonometric substitution, which is one of the toughest integration methods.

Let x = 4/3 tan(t). Then
du = (4/3) sec^2(t) dt, so our integral is now

Integral (1 / (9[16/9]tan^2(t) + 16) ) (4/3) sec^2(t) dt

Skipping some details which I'll leave you to figure out, this will leave you with

(1/12) * Integral (1/(tan^2(t)) (sec^2(t)) dt

By trig identities (and skipping some more details), the stuff inside the integral should reduce to

(1/12) * Integral (sin^2(t)) dt

And, by the half angle identity, sin^2(t) = (1/2)(1 - cos(2t)), so we have

(1/12) * Integral ( (1/2) (1 - cos(2t)) ) dt

Pulling out the constant 1/2,

(1/24) * Integral (1 - cos(2t)) dt

Which we now integrate normally.

(1/24) [t - (1/2)sin(2t)] + C, or

(1/24)t - (1/48)sin(2t) + C

We have to use the double angle identity on sin(2t).

(1/24)t - (1/48) 2sin(t)cos(t) + C
(1/24)t - (1/24) sin(t)cos(t) + C

Recall that we let
x = (4/3) tan(t). This means
tan(t) = (3x)/4

Remember that by SOHCAHTOA, tan(t) = opp/adj
Since tan(t) = (3x)/4, then

opp = 3x
adj = 4
and it follows, by Pythagoras,
hyp = sqrt( [3x]^2 + [4]^2) = sqrt(9x^2 + 16)

Recall that
(1/24)t - (1/24) sin(t)cos(t) + C

sin(t) = opp/hyp = (3x) / sqrt(9x^2 + 16)
cos(t) = adj/hyp = 4/sqrt(9x^2 + 16)

t = arctan([3x]/4), so we have our integral as

(1/24)arctan(3x/4) - (1/24) [(3x) / sqrt(9x^2 + 16)] [4/sqrt(9x^2 + 16)] + C

Reducing this,

(1/24)arctan(3x/4) - (1/24) [ (12x) / (9x^2 + 16) ] + C

And, sure, we can reduce it more.

(1/24)arctan(3x/4) - (1/2) (x/(9x^2 + 16)) + C