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

A function whose derivative is the given function.

The arbitrary constant 'C' added to the antiderivative.

A technique for simplifying integrals by substituting a part of the integrand with a new variable 'u'.

A trinomial that can be factored into the square of a binomial.

The family of all antiderivatives of a function.

A rule to find the antiderivative of a power function.

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

A technique used to rewrite a quadratic expression into a perfect square trinomial plus a constant.

To simplify trigonometric integrals into forms that are easier to evaluate.

$\ln|x| + C$

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

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

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

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

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

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

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

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

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

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