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Common integrals

Integration of a function is computed using the primitive. To check the answer, differentiate the right hand side of the following. $$c$$ is a constant.

Power function For $$\alpha\ne -1,0$$, $$x > 0$$ $$$\int x^\alpha\,dx = \frac{x^{\alpha + 1}}{\alpha + 1} + c,\quad \int x^{-1}\,dx = \ln\vert x\vert + c $$$ The formula extends for $$x\ne0$$, $$\alpha = -1$$ and for $$x\in\R$$, $$\alpha \in \N$$.

Proof We differentiate the right hand side (taking $$c=0$$). $$$\frac{(x^{\alpha+1})'}{\alpha + 1} = x^\alpha, \qquad (\ln\vert x\vert)' = \frac{1}{x} $$$

The two red functions are both primitives of the blue polynomial function.
The two red functions are both primitives of the blue polynomial function.

Integration of a function is computed using the primitive. To check the answer, differentiate the right hand side of the following. $$c$$ is a constant.

  • Exponential function: For $$\alpha>0$$, $$$\int e^x\,dx = e^x + c,\quad \int \alpha ^x\,dx = \frac{\alpha^x}{\ln\alpha } + c $$$
  • Logarithm: $$$\int \ln\vert x\vert\,dx = x\ln\vert x\vert - x +c $$$

Proof We differentiate the right hand side (taking $$c=0$$).

For the exponential function, we have $$$(e^x)' = e^x ,\quad \frac{(\alpha^x)'}{\ln\alpha } = \alpha^x $$$

For the logarithm, we have $$$ (x\ln\vert x\vert - x)' = \ln\vert x\vert + \frac{x}{x} - 1 = \ln\vert x\vert. $$$

The red functions are primitives of the blue.
The red functions are primitives of the blue.

Integration of a function is computed using the primitive. To check the answer, differentiate the right hand side of the following. $$c$$ is a constant.

Trigonometric functions \begin{gather*} \int \sin x\,dx = -\cos x+c, \; \int\cos x\,dx = \sin x+c,\\ \int \tan x\,dx = -\ln\vert \cos x\vert + c,\; \int \cot x\,dx = \ln\vert \sin x\vert + c, \end{gather*}

Proof: We differentiate the right hand side (taking $$c=0$$). \begin{gather*} -(\cos x)' = \sin x,\quad (\sin x)' = \cos x,\\ -(\ln\vert \cos x\vert)' = \frac{-(\cos x)'}{\cos x} = \frac{\sin x}{\cos x} = \tan x\\ (\ln\vert \sin x\vert)' = \frac{(\sin x)'}{\sin x} = \frac{\cos x}{\sin x} = \cot x\\ \end{gather*}

The two red functions are both primitives of the blue function, $$\sin x$$
The two red functions are both primitives of the blue function, $$\sin x$$

For these rational function, $$a$$ must not be $$0$$. The primitives of these rational functions are inverse trigonometric functions.

\begin{gather*} \int \frac{1}{\sqrt{a^2-x^2}}\,dx = \arcsin(x/a) + c = - \arccos(x/a) + c',\\ \int \frac{a}{a^2+x^2}\,dx = \arctan(x/a) + c = - \arccot(x/a) + c'. \end{gather*}

Proof We differentiate the right hand side (taking $$c=0$$). For $$a\ne 0$$, \begin{gather*} \arcsin\big(\frac{x}{a}\big)' = \frac{1}{a\sqrt{1 - (x/a)^2}} = \frac{1}{ \sqrt{a^2 - x^2}}\\ \arctan\big(\frac{x}{a}\big)' = \frac{1}{a\big(1 + (x/a)^2\big)} = \frac{a}{a^2 + x^2} \end{gather*} The formulae with $$\arccos$$ and $$\arccot$$ are obtained either by direct differentiation or by using the identities $$$ \arccos x + \arcsin x =\frac{\pi}{2},\quad \arctan x + \arccot x =\frac{\pi}{2}. $$$