Question

what is the formula of integration 1/x square+a square dx​

Answer 1

Answer:

I=∫1x2–a2dx⇒I=∫1(x–a)(x+a)dx⇒I=12a∫[(x +a)–(x–a)](x–a)(x+a)dx⇒∫dxx2–a2=12a[∫1x –adx–∫1x+adx]. Using the integral formula .

Answer 2

⇒∫(1/x²+a²)dx

⇒here let us take x=atant

⇒dx=asec²tdt

⇒so, x²=a²tant

⇒∫(1/a²tan²t+a²)asec²tdt

⇒∫(asec²t/a²sec²t)dt

⇒∫(1/a)dt

⇒t/a

so, we have taken x=atant

so, t=tan inverse(x/a)

so answer will be 1/a(tan inverse(x/a)).