Question

Integration e^x(tan^-1 x +1/1+x²)

Answer 1

EXPLANATION.

$\implies \displaystyle \int e^{x} \bigg(tan^{-1} x \ + \dfrac{1}{1 + x^{2} } \bigg)dx$

As we know that,

Formula of :

⇒ ∫eˣ[f(x) + f'(x)]dx = eˣ f(x) + c.

As we know that,

Differentiation of tan⁻¹x.

⇒ y = tan⁻¹x.

⇒ dy/dx = 1/1 + x².

Using this formula in the equation, we get.

$\implies \displaystyle \int e^{x} \bigg(tan^{-1} x + \dfrac{1}{1 + x^{2} } \bigg) dx \ = e^{x} tan^{-1} x + C$

                                                                                                                       

MORE INFORMATION.

Standard integrals.

(1) = ∫0.dx = c.

(2) = ∫1.dx = x + c.

(3) = ∫k dx = kx + c, (k ∈ R).

(4) = ∫xⁿdx = xⁿ⁺¹/n + 1 + c, (n ≠ -1).

(5) = ∫dx/x = ㏒(x) + c.

(6) = ∫eˣdx = eˣ + c.

(7) = ∫aˣdx = aˣ/㏒(a) + c = aˣ㏒(e) + c.

Answer 2

Answer:

Answer

Step-by-step explanation:

Just give a thanks ❤️ for answer.