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The integral of $\arctan x$ or the inverse of tan x is the function that returns the inverse tangent of x as its derivative. It is equal to: $\int \arctan x\phantom{x}dx= x \arctan x -\dfrac{1}{2} \ln|1 + x^2| + C$.
In this complete guide, learn how to derive the formula for arctan x and how to apply this to find other integrals. After going through this article, you will feel confident when working with inverse tangent integrals and applications of the arctan integral.
What Is the Integral of Arctan(x)?
The integral of arctan x or the inverse of tan x is equal to $\int \arctan x\phantom{x}dx= x \arctan x -\dfrac{1}{2} \ln|1 + x^2| + C$. From the expression, the integral of arctan (x) results to two expressions: the product of the x and \arctan x and a logarithmic expression $\dfrac{1}{2} \ln|1 + x^2|$.
Constant of Integration
The term $C$ represents the constant of integration, and is often used for the indefinite integral of arctan x.
\begin{aligned}\int \arctan x \phantom{x}dx &= {\color{Purple} x \arctan x } – {\color{Teal} \dfrac{1}{2}|1+x^2|}+{\color{Pink}C}\end{aligned}
The integral of arctan x is the result of applying the integration by parts. You can also find the integrals of inverse trigonometric functions (arcos integral and arcsin integral) from this method. We also use integral by parts to evaluate the hyperbolic functions such as the integral of arctanhx, arcsinhx, and arcoshx. This is why we’ve allotted a special section breaking down the steps for you!
How To Find the Integral of Arctan x
To find the integral of $\arctan x$, apply the integration by parts method. Since $arctan x$ is a single function, rewrite it as a product of $1$ and $\arctan x$ itself. This leads to an expression that is a product of two functions: $u = 1$ and $v = \arctan x$. Take a quick refresher on integration by parts before working on the integral of $\arctan x$:
• After assigning the proper factors to be $u$ and $dv$, find the expressions for $du$ and $v$. Use the table below as a guide.
\begin{aligned}u &= f(x)\end{aligned} | \begin{aligned}dv &= g(x)\phantom{x}dx\end{aligned} |
\begin{aligned}du &= f^{\prime}(x)\phantom{x}dx\end{aligned} | \begin{aligned}v &= \int g(x)\phantom{x}dx\end{aligned} |
• Use the appropriate rules to differentiate and integrate the expressions.
• Apply the integral by parts formula, $\int u \cdot dv = uv – \int v \cdot du$, given that $\int u \phantom{x}dv = \int f(x)g(x) \phantom{x}dx$.
These are the crucial steps to remember when finding the integral of $\arctan x$. In the next section, learn how to apply this method to evaluate the expression for $\arctan x$.
Integration by Parts and Arctan x
When using the integration by parts to find $\arctan x$, it’s important to select the right expression for $u$. This is where the “LIATE” mnemonic comes in. As a refresher, LIATE stands for: Logarithmic, Inverse Logarithmic, Algebraic, Trigonometric, and Exponential. This is the order when prioritizing the factor and assigning the expression for $u$.
For $\int \arctan x\phantom{x} dx =\int \arctan x \cdot 1\phantom{x}dx $, assign $u$ as $\arctan x$ or $\tan^{-1} x$. This also means that $dv $ is equal to $1 \phantom{x}dx$. Now, find the expressions for $du$ and $v$.
• Use the fact that $\dfrac{d}{dx} \arctan x = \dfrac{1}{1+ x^2}$.
• Integrate both sides of the second equation to find $v$.
\begin{aligned}u &=\arctan x\end{aligned} | \begin{aligned}dv &= 1\phantom{x}dx\end{aligned} |
\begin{aligned}du &= \dfrac{1}{1+x^2} \phantom{x}dx\end{aligned} | \begin{aligned}v &=\int 1\phantom{x} dx\\&= x +C\end{aligned} |
We now have all the components to find the integral of $\arctan x$ using integration by parts. So apply the formula $\int u \cdot dv = uv – \int v \cdot du$ as shown below.
\begin{aligned}\int u \cdot dv &= uv – \int v \cdot du \\\int \arctan x \cdot 1 \phantom{x}dx &= x \cdot \arctan x – \int x \cdot \dfrac{1}{1 + x^2}\phantom{x} dx\end{aligned}
Now, apply algebraic and integral techniques to further simplify the second part of the expression in $ x \cdot \arctan x – \int x \cdot \dfrac{1}{1 + x^2}$. This means we’ll disregard $x\arctan x$ for now and focus on $\int \dfrac{x}{1+x^2}\phantom{x}dx$. Rewrite $\int x \cdot \dfrac{1}{1 + x^2}\phantom{x} dx$ by adding $\dfrac{1}{2}$ as an external factor. Multiply the integrand by $2$ to balance this new factor.
\begin{aligned}\int x \cdot \dfrac{1}{1 + x^2} \phantom{x}dx &= \int \dfrac{x}{1 +x^2}\phantom{x}dx\\&= \dfrac{1}{2}\int \dfrac{2x}{1 +x^2}\phantom{x}dx\end{aligned}
Use the u-substitution to evaluate the resulting expression. For the case of $\dfrac{1}{2}\int \dfrac{2x}{1 +x^2}\phantom{x}dx$, use $u = 1+ x^2$ and so, $du = 2x \phantom{x}dx$.
\begin{aligned}u =1+x^2 &\Rightarrow du =2x\phantom{x}dx\\\dfrac{1}{2}\int \dfrac{2x}{1 +x^2}\phantom{x}dx &= \dfrac{1}{2}\int \dfrac{1}{u}\phantom{x}du\\&=\dfrac{1}{2}\ln|u| +C\\&=\dfrac{1}{2}\ln|1 +x^2| +C\end{aligned}
Use this to rewrite the previous expression for $\int \arctan x\phantom{x}dx$.
\begin{aligned}\int \arctan x\phantom{x}dx &=x\arctan x – \dfrac{1}{2}\int \dfrac{2x}{1 + x^2}\phantom{x} dx\\&=x\arctan x – \dfrac{1}{2}\ln|1 +x^2| + C\end{aligned}
This confirms that the integral of $\arctan x$ is equal to $ x\arctan x – \dfrac{1}{2}\ln|1 +x^2| + C$.
There is no need to use this long process when working on other arctan integral examples. All you have to do is use the established formula for $\int \arctan x$ and other simpler integral methods. Don’t worry, you’ll have a chance to work on different examples in the next section!
How to Use the Integral of $\arctan x$ To Evaluate Integrals
Rewrite the affected function so that it is of the form: $\arctan x$.
Use this technique when an integrand contains an inverse trigonometric function. Once in the simplest form, use the formula for the integral of $\arctan x$, $\int \arctan x\phantom{x}dx = x\arctan x – \dfrac{1}{2}\ln|1 +x^2| + C$.
In most cases, you’ll need to use the $u$-substitution method. Here are some steps to follow when using the formula for the integral of $\arctan x$:
• Assign the appropriate term for $u$.
• Rewrite the involved inverse trigonometric function as $\arctan u$.
• Apply the formula for $\int \arctan x\phantom{x}dx$.
You’ll need more algebraic techniques and other integration methods for some instances. But what’s important is that you now know how to find the integrals that involve arctan x. Why don’t you try out the different examples shown below? Test your understanding of arctan x and its integral!
Evaluating the Integral of arctan(4x)
Apply the $u$-substitution to evaluate $\int \arctan 4x\phantom{x} dx$. First, let $u$ represent $4x$, so this leads to $du = 4 \phantom{x}dx$ and $\arctan 4x =\arctan u$. Rewrite the integral as shown below.
\begin{aligned}u =4x &\Rightarrow du =4\phantom{x} dx\\\int \arctan 4x\phantom{x} dx&=\int \arctan u \cdot\dfrac{1}{4} du\\&=\dfrac{1}{4}\int \arctan u\phantom{x}du\end{aligned}
The integral is in the simplest form, $\int \arctan u\phantom{x}du$, so apply the formula for the integral of inverse tangent functions.
\begin{aligned}\dfrac{1}{4}\int \arctan u\phantom{x}du&= \dfrac{1}{4}\left(u\arctan u – \dfrac{1}{2}\ln|1 +u^2| + C\right)\\&=\dfrac{u}{4}\arctan u – \dfrac{1}{8}\ln|1 +u^2| + C\end{aligned}
Rewrite the resulting integral by replacing $u$ back to $4x$. Simplify the resulting expression as shown below.
\begin{aligned}\dfrac{u}{4}\arctan u – \dfrac{1}{8}\ln|1 +u^2| + C&=\dfrac{4x}{4}\arctan 4x – \dfrac{1}{8}\ln|1 +(4x)^2| + C\\&=x\arctan 4x – \dfrac{1}{8}\ln|1 +16x^2| + C\end{aligned}
This shows that the integral of $\arctan 4x$ is equal to $ x\arctan 4x – \dfrac{1}{8}\ln|1 +16x^2| + C$.
Evaluating the Integral of arctan(6x)
Apply a similar process to evaluate $\int \arctan 6x \phantom{x}dx$. Use the $u$-substitution and let $u$ be equal to $6x$. This simplifies the integral expression to $\int \arctan u \phantom{x}du$. Find the integral using the formula $\int \arctan x\phantom{x}dx = x\arctan x – \dfrac{1}{2}\ln|1 +x^2| + C$.
\begin{aligned}u =6x &\Rightarrow du = 6\phantom{x}dx\\\int \arctan 6x \phantom{x}dx&= \dfrac{1}{6}\int\arctan u \phantom{x}du\\&=\dfrac{1}{6}\left(u\arctan u – \dfrac{1}{2}\ln|1 +u^2| + C\right)\\&=\dfrac{u}{6}\arctan u -\dfrac{1}{12}\ln|1 +u^2|+C\end{aligned}
Replace $u$ with $6x$ then simplify the resulting expression.
\begin{aligned}\dfrac{u}{6}\arctan u -\dfrac{1}{12}\ln|1 +u^2|+C&= \dfrac{6x}{6}\arctan 6x -\dfrac{1}{12}\ln|1 +(6x)^2|+C\\&=x\arctan 6x -\dfrac{1}{12}\ln|1 +36x^2|+C\end{aligned}
This shows that $\int \arctan 6x \phantom{x}dx = x\arctan 6x -\dfrac{1}{12}\ln|1 +36x^2|+C$.
Evaluating the definite integral $\int_{0}^{1} \arctan \dfrac{x}{2}\phantom{x}dx$
When evaluating definite integrals involving $\arctan x$, use the same process. But this time, evaluate the resulting expression at lower and upper limits. For $\int_{0}^{1} \arctan \dfrac{x}{2}\phantom{x}dx$, focus on evaluating the integral as if it’s an indefinite integral. Use the $u$-substitution method as we have applied it in the previous problems.
\begin{aligned}u = \dfrac{x}{2} &\Rightarrow du = \dfrac{1}{2}\phantom{x}dx\\\int\arctan \dfrac{x}{2}\phantom{x}dx&= 2\int\arctan u\phantom{x}du\\&=2(u\arctan u – \dfrac{1}{2}\ln|1 +u^2|) + C\\&=2\left[\dfrac{x}{2}\arctan\dfrac{x}{2} – \dfrac{1}{2}\ln\left|1 +\left(\dfrac{x}{2}\right)^2\right|\right] + C\\&= x\arctan \dfrac{x}{2} -\ln \left|1 +\dfrac{x^2}{4}\right| + C\end{aligned}
Now, evaluate this resulting expression from $x=0$ to $x=1$ to find the definite integral’s value.
\begin{aligned}\int_{0}^{1} \arctan \dfrac{x}{2}\phantom{x}dx &=\left[x\arctan \dfrac{x}{2} -\ln \left|1 +\dfrac{x^2}{4}\right|\right]_{\displaystyle{0}}^{\displaystyle{1}}\\&=\left(1\arctan \dfrac{1}{2} – \ln\left|1+\dfrac{1}{4}\right|\right)-\left(0\arctan 0 – \ln\left|1+0\right|\right)\\&=\arctan\dfrac{1}{2} -\ln\dfrac{5}{4}\end{aligned}
Hence, $\int_{0}^{1} \arctan \dfrac{x}{2}\phantom{x}dx = \arctan\dfrac{1}{2} -\ln\dfrac{5}{4}$.