Evaluate the definite integral arctan X dx
The Correct Answer and Explanation is:
To evaluate the definite integral of arctan(x)\arctan(x)arctan(x), we can use integration by parts. The general formula for integration by parts is:∫u dv=uv−∫v du\int u \, dv = uv – \int v \, du∫udv=uv−∫vdu
Step 1: Choose uuu and dvdvdv
For the integral of arctan(x)\arctan(x)arctan(x), we choose:
- u=arctan(x)u = \arctan(x)u=arctan(x) (because it simplifies when differentiated)
- dv=dxdv = dxdv=dx
Step 2: Differentiate uuu and integrate dvdvdv
- The derivative of u=arctan(x)u = \arctan(x)u=arctan(x) is du=11+x2 dxdu = \frac{1}{1 + x^2} \, dxdu=1+x21dx
- The integral of dv=dxdv = dxdv=dx is v=xv = xv=x
Step 3: Apply the integration by parts formula
Now, applying the integration by parts formula:∫arctan(x) dx=xarctan(x)−∫x1+x2 dx\int \arctan(x) \, dx = x \arctan(x) – \int \frac{x}{1 + x^2} \, dx∫arctan(x)dx=xarctan(x)−∫1+x2xdx
Step 4: Simplify the second integral
We now need to evaluate the integral:∫x1+x2 dx\int \frac{x}{1 + x^2} \, dx∫1+x2xdx
Notice that the numerator xxx is the derivative of 1+x21 + x^21+x2, so this is a standard integral:∫x1+x2 dx=12ln(1+x2)\int \frac{x}{1 + x^2} \, dx = \frac{1}{2} \ln(1 + x^2)∫1+x2xdx=21ln(1+x2)
Step 5: Final expression
Now substitute this result back into the integration by parts formula:∫arctan(x) dx=xarctan(x)−12ln(1+x2)+C\int \arctan(x) \, dx = x \arctan(x) – \frac{1}{2} \ln(1 + x^2) + C∫arctan(x)dx=xarctan(x)−21ln(1+x2)+C
Explanation:
The method of integration by parts is useful when dealing with products of functions, especially when one of the functions becomes simpler when differentiated. In this case, the arctangent function is simplified by differentiating, and the resulting expression leads to an easily solvable integral. The final result is the combination of the product term xarctan(x)x \arctan(x)xarctan(x) and a logarithmic term derived from the simpler integral.
Therefore, the indefinite integral of arctan(x)\arctan(x)arctan(x) is:∫arctan(x) dx=xarctan(x)−12ln(1+x2)+C\int \arctan(x) \, dx = x \arctan(x) – \frac{1}{2} \ln(1 + x^2) + C∫arctan(x)dx=xarctan(x)−21ln(1+x2)+C
This result can be used to evaluate definite integrals by substituting the limits into the antiderivative expression.
