Integral Of X Arctan X: A Better Way To Teach It
The integral of x arctan x is $$\int x\arctan(x)\,dx = \frac{x^2}{2}\arctan(x) - \frac{x}{2} + \frac{1}{2}\arctan(x) - \frac{1}{2}\ln(1+x^2) + C$$, and the cleanest method is integration by parts. MIT OpenCourseWare places this topic within its "Techniques of Integration" unit, which is the right framework for this kind of product integral.
Core Result
The standard antiderivative can also be written in a compact factored form as $$\frac{1}{2}(x^2+1)\arctan(x) - \frac{x}{2} - \frac{1}{2}\ln(1+x^2) + C$$. The two forms are equivalent, and both are useful depending on whether you want an expanded answer or a presentation-friendly expression.
Purposeful Steps
Use integration by parts with $$u=\arctan(x)$$ and $$dv=x\,dx$$, because the inverse tangent becomes simpler when differentiated and the polynomial becomes simpler when integrated.
- Set $$u=\arctan(x)$$ and $$dv=x\,dx$$.
- Then $$du=\frac{1}{1+x^2}\,dx$$ and $$v=\frac{x^2}{2}$$.
- Apply $$\int u\,dv = uv-\int v\,du$$.
- Substitute to get $$\int x\arctan(x)\,dx = \frac{x^2}{2}\arctan(x)-\frac{1}{2}\int \frac{x^2}{1+x^2}\,dx$$.
- Rewrite $$\frac{x^2}{1+x^2}=1-\frac{1}{1+x^2}$$, then integrate term by term.
Step-by-Step Derivation
After the rewrite, the remaining integral becomes $$\frac{1}{2}\int \left(1-\frac{1}{1+x^2}\right)dx$$, which splits into a linear term and an arctangent term. This gives $$\frac{1}{2}\left(x-\arctan(x)\right)$$, so the full antiderivative is $$\frac{x^2}{2}\arctan(x)-\frac{x}{2}+\frac{1}{2}\arctan(x)-\frac{1}{2}\ln(1+x^2)+C$$.
"Choose the part that simplifies most when differentiated, and let the other part integrate easily."
Why It Works
The key idea in integration by parts is that product integrals often become easier when one factor is differentiated and the other is integrated. In this problem, $$\arctan(x)$$ differentiates to $$\frac{1}{1+x^2}$$, which pairs neatly with the algebraic rewrite $$\frac{x^2}{1+x^2}=1-\frac{1}{1+x^2}$$.
| Component | Expression | Role in the solution |
|---|---|---|
| $$u$$ | $$\arctan(x)$$ | Differentiates to a simpler rational function. |
| $$dv$$ | $$x\,dx$$ | Integrates to $$\frac{x^2}{2}$$. |
| $$du$$ | $$\frac{1}{1+x^2}\,dx$$ | Creates the logarithm after simplification. |
| Final antiderivative | $$\frac{x^2}{2}\arctan(x)-\frac{x}{2}+\frac{1}{2}\arctan(x)-\frac{1}{2}\ln(1+x^2)+C$$ | Equivalent to the compact form. |
Common Errors
- Using $$u=x$$ and $$dv=\arctan(x)\,dx$$, which makes the integral harder instead of easier.
- Forgetting the product rule structure in integration by parts.
- Stopping after the first step and missing the algebraic rewrite of $$\frac{x^2}{1+x^2}$$.
- Dropping the constant of integration $$C$$.
Quick Check
Differentiate the result to verify it returns $$x\arctan(x)$$, which is the fastest way to confirm correctness. This verification step is especially useful in calculus because the derivative of $$\ln(1+x^2)$$ is $$\frac{2x}{1+x^2}$$, which cancels exactly with the rearranged middle term.
Learning Context
This integral appears in standard undergraduate calculus courses under techniques of integration, including the MIT OpenCourseWare single-variable calculus sequence. For students and educators, it is a good example of how symbolic manipulation, verification, and method selection work together in rigorous problem solving.
Expert answers to Integral Of X Arctan X A Better Way To Teach It queries
What is the final answer?
$$\int x\arctan(x)\,dx = \frac{x^2}{2}\arctan(x) - \frac{x}{2} + \frac{1}{2}\arctan(x) - \frac{1}{2}\ln(1+x^2) + C$$.
Why use integration by parts?
Because the product $$x\arctan(x)$$ becomes simpler when the inverse tangent is differentiated and the polynomial is integrated, which is exactly the pattern integration by parts is designed for.
Is there a shorter form?
Yes, the same result can be written as $$\frac{1}{2}(x^2+1)\arctan(x)-\frac{x}{2}-\frac{1}{2}\ln(1+x^2)+C$$, which is algebraically equivalent and often easier to present neatly.
