what is the integral of 1/(7+y^2) dy?

Question

step by step, please.

i came up with an answer ln(7+y^2)/2, but it's wrong.
help would be greatly appreciated!

Answer 1

I = ∫ dy / (a ² + y ²) = (1/a) tan^(-1) (y / a) + C
Therefore:-
I = ∫ dy / [ (√7) ² + y ²) ]
I = (1 / √7) tan^(-1)(y / √7) + C

Answer 2

Integral ( 1/(7 + y^2) dy )

To solve this, you have to use trigonometric substitution.

Let y = sqrt(7) tan(t)
dy = sqrt(7) sec^2(t) dt

Integral ( 1/(7 + 7tan^2(t)) sqrt(7) sec^2(t) dt )

Factor out the constant sqrt(7), to get

sqrt(7) * Integral ( 1/(7 + 7tan^2(t)) sec^2(t) dt )

Factor 7 from the denominator of the fraction, which will in turn factor out as (1/7) from the integral, to get

sqrt(7) (1/7) * Integral ( 1/(1 + tan^2(t)) sec^2(t) dt )

Use the identity 1 + tan^2(t) = sec^2(t), to obtain

sqrt(7)/7 * Integral ( 1/sec^2(t) sec^2(t) dt )

Notice how the secants cancel out.

sqrt(7)/7 * Integral ( 1 dt )

Which becomes

(sqrt(7)/7)t + C

But y = sqrt(7) tan(t), so
tan(t) = ( 1/sqrt(7) )y, meaning

t = arctan ( 1/sqrt(7) )y)

Which makes our final answer

(sqrt(7)/7)arctan ( 1/sqrt(7) )y) + C