Substitute part of the integrand with 'u', find du, rewrite the integral in terms of 'u', integrate, and substitute back to 'x'.

Represents the constant of integration because the derivative of a constant is zero.

When integrating rational functions where the degree of the numerator is greater than or equal to the denominator.

A method to rewrite a quadratic expression into a perfect square trinomial plus a constant, facilitating integration.

They simplify complex trigonometric expressions into forms that are easier to integrate.

When the integrand contains a composite function and its derivative (or a multiple thereof).

Choose a suitable 'u' that simplifies the integral.

To choose the correct method to find the antiderivative of a given expression.

$\arcsin(x) + C$

$\arctan(x) + C$

Apply the power rule: $\frac{x^{5+1}}{5+1} + C = \frac{x^6}{6} + C$

Use u-substitution: let $u = 2x$, then $du = 2dx$. So, $\frac{1}{2} \int \cos(u) du = \frac{1}{2} \sin(u) + C = \frac{1}{2} \sin(2x) + C$

Use u-substitution: let $u = x^2 + 1$, then $du = 2x dx$. So, $\frac{1}{2} \int \frac{1}{u} du = \frac{1}{2} \ln|u| + C = \frac{1}{2} \ln|x^2 + 1| + C$

Use u-substitution: let $u = 3x$, then $du = 3dx$. So, $\frac{1}{3} \int e^u du = \frac{1}{3} e^u + C = \frac{1}{3} e^{3x} + C$

Expand and use power rule: $\int (x^2 + 2x + 1) dx = \frac{x^3}{3} + x^2 + x + C$

Use u-substitution: let $u = x+2$, then $du = dx$. So, $\int \frac{1}{u} du = \ln|u| + C = \ln|x+2| + C$

Use inverse tangent: $\frac{1}{2} \arctan(\frac{x}{2}) + C$

Perform long division: $\int (x+3) dx = \frac{x^2}{2} + 3x + C$

Use the identity $\sin^2(x) = \frac{1 - \cos(2x)}{2}$. Then, $\int \frac{1 - \cos(2x)}{2} dx = \frac{x}{2} - \frac{\sin(2x)}{4} + C$

Use inverse sine: $\arcsin(\frac{x}{3}) + C$

$\int x^n dx = \frac{x^{n+1}}{n+1} + C$, where $n \neq -1$

$\int \sin(x) dx = -\cos(x) + C$

$\int \cos(x) dx = \sin(x) + C$

$\int \tan(x) dx = -\ln|\cos(x)| + C$

$\int \cot(x) dx = \ln|\sin(x)| + C$

$\int \sec(x) dx = \ln|\sec(x) + \tan(x)| + C$

$\int \csc(x) dx = -\ln|\csc(x) + \cot(x)| + C$

$\int e^x dx = e^x + C$

$\frac{d}{dx}(\sin^{-1}(x)) = \frac{1}{\sqrt{1-x^2}}$

$\frac{d}{dx}(\tan^{-1}(x)) = \frac{1}{1+x^2}$