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Question Number 46856 by maxmathsup by imad last updated on 01/Nov/18

$$findf\left(t\right)={\int }_0^1{x}^{2}arctan(1+tx)dx$$

Commented by maxmathsup by imad last updated on 02/Nov/18

$$changementtx=ugivef\left(t\right)={\int }_0^t\frac{u^2}{t^2}arctan(1+u)\frac{du}{t}$$$$=\frac{1}{t^3}{\int }_0^t{u}^{2}arctan(1+u)du(wesupposeu\ne 0)but$$$${\int }_0^t{u}^{2}arctan(1+u)du{=}_{1+u=\alpha }{\int }_1^{1+t}{(\alpha -1)}^{2}arctan\alpha d\alpha$$$$={\left[\frac{1}{3}{(\alpha -1)}^{3}arctan\alpha \right]}_1^{1+t}-\frac{1}{3}{\int }_1^{1+t}{(\alpha -1)}^3\frac{d\alpha }{1+{\alpha }^2}$$$$=\frac{1}{3}\left\{t^{3}arctan(1+t)\right\}-\frac{1}{3}{\int }_1^{1+t}\frac{{\alpha }^3-3{\alpha }^2+3\alpha -1}{1+{\alpha }^2}d\alpha$$$${\int }_1^{1+t}\frac{{\alpha }^3-3{\alpha }^2+3\alpha -1}{1+{\alpha }^2}d\alpha ={\int }_1^{1+t}\frac{\alpha ({\alpha }^2+1)+2\alpha -3{\alpha }^2-1}{1+{\alpha }^2}d\alpha$$$$={\int }_1^{1+t}\alpha d\alpha +{\int }_1^{1+t}\frac{2\alpha }{1+{\alpha }^2}d\alpha -{\int }_1^{1+\alpha }\frac{3{\alpha }^2+1}{1+{\alpha }^2}d\alpha$$$$={\left[\frac{{\alpha }^2}{2}\right]}_1^{1+t}+{\left[ln(1+{\alpha }^2)\right]}_1^{1+t}-3{\int }_1^{1+t}\frac{1+{\alpha }^2-1}{1+{\alpha }^2}d\alpha -{\left[arctan\alpha \right]}_1^{1+t}$$$$=\frac{1}{2}\{{(1+t)}^2-1\}+ln(1+{(1+t)}^2)-ln\left(2\right)-3t+{\left[2arctan\alpha \right]}_1^{1+t}$$$$=\frac{1}{2}\{t^2+2t\}+ln(t^2+2t+2)-ln\left(2\right)-3t+2\{artan(1+t)-\frac{\pi }{2}\}.$$

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