H2 Maths Formulas, Techniques & Graphs >> Calculus >> Integration techniques >>

Integrate Algebraic & Exponential Terms

Constants and single terms of $x$

  
    $$f(x)$$
    $$\int f(x) \phantom{.} dx $$
    Remarks
  

    $$a$$
    $$ax + C$$
    $a$ is a constant
  

    $$a x^n$$
    $${a \over n + 1} x^{n+ 1} + C$$
    $a$ is a constant and $n \ne -1$
  

    $$a x^{-1}= {a \over x}$$
    $$a \ln |x| + C$$
    $a$ is a constant
  

$$f(x)$$	$$\int f(x) \phantom{.} dx $$	Remarks
$$a$$	$$ax + C$$	$a$ is a constant
$$a x^n$$	$${a \over n + 1} x^{n+ 1} + C$$	$a$ is a constant and $n \ne -1$
$$a x^{-1}= {a \over x}$$	$$a \ln \|x\| + C$$	$a$ is a constant

Integrating terms in the form of $ f'(x) [f(x)]^n $ & ${f'(x) \over f(x)}$

For $n \ne -1$, $$ \boxed{ \int f'(x) [f(x)]^n \phantom{.} dx = { [f(x)]^{n + 1} \over n + 1} + C } $$

For $n = -1$, $$ \boxed{ \int f'(x) [f(x)]^{-1} \phantom{.} dx = \int {f'(x) \over f(x)} \phantom{.} dx = \ln \left| f(x) \right| + C } $$

An example using the first formula: \begin{align} \int {x + 1 \over (x^2 + 2x - 5)^2} \phantom{.} dx & = {1 \over 2} \int {2x + 2 \over (x^2 + 2x - 5)^2} \phantom{.} dx \\ & = {1 \over 2} \int (2x + 2) (x^2 + 2x - 5)^{-2} \phantom{.} dx \\ & = {1 \over 2} \left[ (x^2 + 2x - 5)^{-1} \over -1 \right] \\ & = -{1 \over 2(x^2 + 2x - 5)} + C \end{align}

An example using the second formula: \begin{align} \int {4x + 4 \over x^2 + 2x - 5} \phantom{.} dx & = 2 \int {2x + 2 \over x^2 + 2x - 5} \phantom{.} dx \\ & = 2 \ln |x^2 + 2x - 5| + C \end{align}

Exponential terms

  
    $$f(x)$$
    $$\int f(x) \phantom{.} dx $$
  
    $$ f'(x). e^{f(x)} $$
    $$ e^{f(x)} $$

$$f(x)$$	$$\int f(x) \phantom{.} dx $$
$$ f'(x). e^{f(x)} $$	$$ e^{f(x)} $$

Example: \begin{align} \int \sin x \phantom{.} e^{\cos x} \phantom{.} dx & = - \int - \sin x \phantom{.} e^{\cos x} \phantom{.} dx \\ & = -e^{\cos x} + C \end{align}

The formulas can be used together with the formulas in the previous section. For example, \begin{align} \int {e^{3x} \over 1 - e^{3x} } \phantom{.} dx & = -{1 \over 3} \int { -3e^{3x} \over 1 - e^{3x} } \phantom{.} dx \phantom{00000} \left[ \int {f'(x) \over f(x)} \phantom{.} dx \right] \\ & = -{1 \over 3} \ln |1 - e^{3x}| + C \end{align}