Properties of indefinite integrals
Theorem (Linearity property of indefinite integrals)
Let $\mathrm{f}$ and $\mathrm{g}$ be functions, and $a, b$ constants. Then the following hold: \[
\displaystyle\int \big(a \mathrm{f}(x) + b \mathrm{g}(x) \big) \, \mathrm{d}x =
a \displaystyle\int \mathrm{f}(x) \, \mathrm{d}x + b \displaystyle\int \mathrm{g}(x) \, \mathrm{d}x
\]
Proof.
Differentiate the expression \[
a \displaystyle\int \mathrm{f}(x) \, {\rm d}x + b \displaystyle\int \mathrm{g}(x) \, {\rm d}x
\] with respect to $x$ yields: \[
\begin{align*}
 \frac{\\mathrm{d}}{\\mathrm{d}x} \big( a \displaystyle\int \mathrm{f}(x) \, {\mathrm d}x + b \displaystyle\int \mathrm{g}(x) \, \mathrm{d}x \big)
& = a \big( \frac{\\mathrm{d}}{\\mathrm{d}x} \displaystyle\int \mathrm{f}(x) \, {\mathrm d}x \big) + b \big( \frac{\\mathrm{d}}{\\mathrm{d}x} \displaystyle\int \mathrm{g}(x) \, {\mathrm d}x \big) \\
& = a \mathrm{f}(x) + b \mathrm{g}(x).
\end{align*}
\] Hence, $\displaystyle a \displaystyle\int \mathrm{f}(x) \, {\mathrm d}x + b \displaystyle\int \mathrm{g}(x) \, {\mathrm d}x$ is an antiderivative of $a \mathrm{f}(x) + b \mathrm{g}(x)$. 
$\Box$
Corollary (Scalar multiple)
Let $\mathrm{f}$ be a function and $k$ a constant. Then \[
\displaystyle\int k \mathrm{f}(x) \, \mathrm{d}x = k \displaystyle\int \mathrm{f}(x) \, \mathrm{d}x.
\]
Proof. Applying the above theorem by setting $a = k$ and $b=0$.
$\Box$

**Theorem** (Linearity property of indefinite integrals)

Let $\mathrm{f}$ and $\mathrm{g}$ be functions, and $a, b$ constants.  Then the following hold: 
\[
\displaystyle \displaystyle\int \big(a \mathrm{f}(x) + b \mathrm{g}(x) \big) \, \mathrm{d}x =
a \displaystyle \displaystyle\int \mathrm{f}(x) \, \mathrm{d}x + b \displaystyle \displaystyle\int \mathrm{g}(x) \, \mathrm{d}x
\]

Differentiate the expression 
\[
a \displaystyle \displaystyle\int \mathrm{f}(x) \, {\rm d}x + b \displaystyle \displaystyle\int \mathrm{g}(x) \, {\rm d}x
\]
with respect to $x$ yields:
\[
\begin{align*}
        \frac{\mathrm{d}}{\mathrm{d}x} \big( a \displaystyle \displaystyle\int \mathrm{f}(x) \, {\mathrm d}x + b \displaystyle \displaystyle\int \mathrm{g}(x) \, \mathrm{d}x \big)
& =  a \big( \frac{\mathrm{d}}{\mathrm{d}x}  \displaystyle \displaystyle\int \mathrm{f}(x) \, {\mathrm d}x \big) + b \big( \frac{\mathrm{d}}{\mathrm{d}x} \displaystyle \displaystyle\int \mathrm{g}(x) \, {\mathrm d}x \big) \\
& = a \mathrm{f}(x) + b \mathrm{g}(x).
\end{align*}
\]
Hence, $\displaystyle a \displaystyle \displaystyle\int \mathrm{f}(x) \, {\mathrm d}x + b \displaystyle \displaystyle\int \mathrm{g}(x) \, {\mathrm d}x$ is an antiderivative of $a \mathrm{f}(x) + b \mathrm{g}(x)$. 

Let $\mathrm{f}$ be a function and $k$ a constant.  Then
\[
\displaystyle \displaystyle\int k \mathrm{f}(x) \, \mathrm{d}x = k \displaystyle \displaystyle\int \mathrm{f}(x) \, \mathrm{d}x.
\]

Techniques of Integration - Techniques of Integration - General

Techniques of Integration - Techniques of Integration - General