4. Integration by Parts

We apply integration by parts with \[\begin{array}{ll} u=\arctan x & dv=3x^2\,dx \\ du=\dfrac{1}{1+x^2}\,dx \quad & v=x^3 \end{array}\] So: \[\begin{aligned} \int_0^1 3x^2\arctan x\,dx &=\left[x^3\arctan x\right]_0^1-\int_0^1 \dfrac{x^3}{1+x^2}\,dx \\ &=1^3\arctan 1-0^3\arctan 0-\int_0^1 \dfrac{x^3}{1+x^2}\,dx \end{aligned}\] For the integrated part, recall \(\arctan 1=\dfrac{\pi}{4}\) and \(\arctan 0=0\). For the remaining integral, we substitute \(z=1+x^2\). Then \(dz=2x\,dx\) and so \(x\,dx=\dfrac{1}{2}\,dz\) and \(x^2=z-1\). We also change limits. Thus: \[\begin{aligned} \int_0^1 3x^2\arctan x\,dx &=\dfrac{\pi}{4}-\int_1^2 \dfrac{z-1}{z}\dfrac{1}{2}\,dz\ \\ &=\dfrac{\pi}{4}-\dfrac{1}{2}\int_1^2 \left(1-\dfrac{1}{z}\right)\,dz \\ &=\dfrac{\pi}{4}-\dfrac{1}{2}\left[\dfrac{}{}z-\ln|z|\right]_1^2 \\ &=\dfrac{\pi}{4}-\dfrac{1}{2}\left[2-\ln(2)\right]+\dfrac{1}{2}[1] \\ &=\dfrac{\pi}{4}-\dfrac{1}{2}+\dfrac{1}{2}\ln(2) \end{aligned}\] Of course, you can always compute a definite integral by first finding the indefinite integral and then evaluating at the limits. \[\begin{aligned} \int 3x^2\arctan x\,dx &=x^3\arctan x-\dfrac{1}{2}(1+x^2)+\dfrac{1}{2}\ln(1+x^2)+C \\ \int_0^1 3x^2\arctan x\,dx &=\left[x^3\arctan x-\dfrac{1}{2}x^2+\dfrac{1}{2}\ln(1+x^2)\right]_0^1 \\ &=\dfrac{\pi}{4}-\dfrac{1}{2}+\dfrac{1}{2}\ln(2) \end{aligned}\]