Question

Integration of dx/a^2+x^2 is ​

Answer 1

Answer:

its integration is

1/a×tan^-1(x/a)+C

Step-by-step explanation:

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Answer 2

Answer:

1/a tan⁻¹ ( x / a )

Step-by-step explanation:

To find ---> ∫ dx / ( x² + a² )

Solution--->

I = ∫ dx / ( x² + a² )

We do it by substitution method , for this

Let,

x = a tanθ

Differentiting with respect to x both sides, we get,

=> dx = a Sec²θ

Now, x = a tanθ

=> x / a = tanθ

Taking tan⁻¹ both sides we get,

=> tan⁻¹ ( x / a ) = θ

Now,

I = ∫ a Sec²θ dθ / { a² + ( a tanθ )² }

= ∫ a Sec²θ dθ / ( a² + a² tan²θ )

We know that, 1 + tan²θ = Sec²θ , using it here , we get,

= ∫ a Sec²θ dθ / a² ( 1 + tan²θ )

= ∫ Sec²θ dθ / a Sec²θ

Sec²θ , cancel , out from numerator and denominator

= ( 1 / a ) ∫ 1 dθ

= ( 1 / a ) θ + C

= ( 1 / a ) tan⁻¹ ( x / a ) + C

Additional information--->

1) ∫ dx / ( x² - a² ) = (1/2a) log (x -a)/(x + a) + C

2) ∫ dx / ( a² - x² ) = 1 / 2a log (a+x)/(a-x ) + C

3)∫ dx / √(a² - x²) = Sin⁻¹ ( x / a ) + C

4)∫dx/√(x²+a²) = log{x + √(x²+a²)} + C

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