Problem of the week 1 - integrals

[Math Tutor]

Week 1. Calculate the definite integral [tex]\int_{1}^{3}\frac{\sqrt[3]{\arctan (x+2)}}{\sqrt[3]{\arctan (x+2)}+\sqrt[3]{\arctan (6-x)}}dx[/tex]

[dduclam]

Week 1. Calculate the definite integral [tex]I=\int_{1}^{3}\frac{\sqrt[3]{\arctan (x+2)}}{\sqrt[3]{\arctan (x+2)}+\sqrt[3]{\arctan (6-x)}}dx[/tex]

My solution for Problem of the week 1:

Put [tex]t=4-x => dt=-dx ; x=1 -> t=3 , x=3 -> t=1[/tex]

[tex]I=\int_{3}^{1}\frac{\sqrt[3]{\arctan (6-t)}}{\sqrt[3]{\arctan (6-t)}+\sqrt[3]{\arctan (2+t)}}(-dt)[/tex]

[tex]=\int_{1}^{3}\frac{\sqrt[3]{\arctan (6-x)}}{\sqrt[3]{\arctan (x+2)}+\sqrt[3]{\arctan (6-x)}}dx[/tex]

=> [tex]2I=I+I=\int_{1}^{3}\frac{\sqrt[3]{\arctan (x+2)}+\sqrt[3]{\arctan (6-x)}}{\sqrt[3]{\arctan (x+2)}+\sqrt[3]{\arctan (6-x)}}dx[/tex]

[tex]=\int_{1}^{3}1.dx=x|^3_1=2[/tex]

Thus [tex]I=1[/tex]


PS: A nother integral problem,nice but not hard:

[tex]J=\int \frac1{1+x^4} dx[/tex]


@teacher: What's the condition of a problem which is problem of week?