The integral of the exponential function , $e^x$, can be found using basic calculus principles. The integral of $e^x$ with respect to (x) is: \[\int e^x \, dx = e^x + C\] Here, (C) represents the constant of integration, which is added because the process of integration determines the antiderivative up to an arbitrary constant. […]

Integration of root a^2 +x^2, x^2 -a^2, a^2 -x^2 is given as $ \int \sqrt {a^2 – x^2} dx = \frac {1}{2} x \sqrt {a^2 – x^2} + \frac {1}{2} a^2 \sin^{-1} \frac {x}{a} + C$ $ \int \sqrt {a^2 + x^2} dx = \frac {1}{2} x \sqrt {a^2 + x^2} + \frac {1}{2} a^2

Integration of cot inverse x can be calculated using integration by parts .Here is the formula for it \[ \int \cot^{-1}x \, dx = x \cot^{-1}x + \frac {1}{2} \ln (1 + x^2) + C \] where (C) is the constant of integration. Proof of integration of cot inverse x To find the integral we

Integration of sec inverse x can be calculated using integration by parts ,integration by substitution .Here is the formula for it \[ \int \sec^{-1}x \, dx = x \sec^{-1}x – \ln |x + \sqrt { x^2- 1}| + C \] where (C) is the constant of integration. Proof of integration of sec inverse x We

Integration of cosec inverse x can be calculated using integration by parts ,integration by substitution .Here is the formula for it \[ \int \csc^{-1}x \, dx = x \csc^{-1}x + \ln |x + \sqrt { x^2+ 1}| + C \] where (C) is the constant of integration. Proof of integration of cosec inverse x We

Integration of 1/a^2+ x^2, 1/a^2- x^2, 1/x^2 -a^2 can be obtained using trigonometric substitution, integration by partial fractions. The formula given are $\int \frac {1}{x^2 + a^2} dx = \frac {1}{a} \tan ^{-1} (\frac {x}{a}) + C$ $\int \frac {1}{x^2 – a^2} dx = \frac {1}{2a} ln |\frac {x-a}{x+a}| + C$ $\int \frac {1}{a^2 –

Integration of rational functions, which are quotients of polynomials, is a fundamental concept in calculus. The general form of a rational function is: $$ \frac{P(x)}{Q(x)} $$ where ( P(x) ) and ( Q(x) ) are polynomials. The strategy for integrating such functions typically involves a few steps: (1) Polynomial Division: If the degree of (

Integration of root tan x is little complex. This can be solve using integration by substitution method . The formula is given by $\int \sqrt {tan x} \; dx = \frac {1}{\sqrt 2} \tan^{-1} (\frac {tan x -1}{\sqrt {2tan x}}) + \frac {1}{2\sqrt 2} \ln \frac {(tan x + 1 -\sqrt {2tan x})}{(tan x +