# Evaluate $\int_{0}^{1} \sqrt{1+\sqrt[3] {x}}\,dx,$

Evaluate $$\int_{0}^{1} \sqrt{1+\sqrt[3] {x}} \,dx,$$

So,i've tried doing the substitution $$u=\sqrt[3] {x}$$( this was a suggestion by wolfram), but frankly i am reaching a point where it doesn't work . Could help me out ?

• I'd have tried $u = 1+\sqrt[3]x$, which gives $dx=3\sqrt[3] {x^2}du= 3(u-1)^2du$
– lulu
Mar 22 '20 at 16:20
• Try $u = 1+\sqrt[3] {x}$ Mar 22 '20 at 16:21

If $$u = \sqrt{1+\sqrt[3]{x}}$$, then the integral becomes $$\int_0^1 \sqrt{1+\sqrt[3]{x}} \ dx = 6 \int_1^\sqrt{2} u^2 \, (u^2-1)^2du.$$ Now, expand the integrand and integrate a polynomial.

Hint:

Let $$u = 1 + \sqrt[3]x\implies\mathrm dx = 3x^{\frac23}\mathrm du$$. Therefore,

$$\int_0^1\sqrt{1 + \sqrt[3]x}\,\mathrm dx\equiv3\int_1^2(u - 1)^2\sqrt u\,\mathrm du = 3\int_1^2 u^{\frac52}\,\mathrm du - 6\int_1^2 u^{\frac32}\,\mathrm du + 3 \int_1^2 u^{\frac 12}\,\mathrm du.$$

Solve the three integrals and plug-in limits.

Hint:

Let $$\sqrt{1+\sqrt[3]x}= y\implies x=((y^2-1)^3, dx=?$$

If $$x=0,y=1; x=1, y=?$$