<s>[INST] The value of the integral $\int_0^2\left(\sqrt{1+x^3}+\sqrt[3]{x^2+2x}\:\right)dx$ The value of definite integral $$\int\limits_{0}^{2}\left(\sqrt{1+x^3}+\sqrt[3]{x^2+2x}\:\right)dx$$ is $$(A)\,4 \quad(B)\,5 \quad (C)\,6  \quad(D)\,7$$
My attempt:
I tried using $\int\limits_{0}^{a}f(x)dx=\int\limits_{0}^{a}f(a-x)dx$ but not working. I tried putting $x^3+1=\tan^2\theta$, its also not working.
Can someone help me solve this problem? 
 [/INST] I have one more way to do this:
Let $$I_1=\int_{0}^{2} \sqrt{x^3+1} \:dx$$ By Integration by parts we get
$$I_1=x\sqrt{x^3+1} \bigg|_{0}^{2}-\int_{0}^{2}\frac{3x^3 dx}{2\sqrt{x^3+1}} \tag{1}$$
Now let $$I_2=\int_{0}^{2} (x^2+2x)^{\frac{1}{3}}dx$$ Use substitution $x^2+2x=t^3$ we get
$(2x+2)dx=3t^2dt$ that is
$dx=\frac{3t^2dt}{2(x+1)}$
But $(x+1)^2-1=t^3$ $\implies$
$x+1=\sqrt{t^3+1}$
Also the limits will be again $0$ and $2$
Thus
$$I_2=\int_{0}^{2}\frac{3t^3dt}{2(x+1)}=\int_{0}^{2}\frac{3t^3dt}{2\sqrt{t^3+1}}$$  since $t$ is Dummy variable
$$I_2=\int_{0}^{2}\frac{3x^3dx}{2\sqrt{x^3+1}} \tag{2}$$
Adding $(1)$ and $(2)$ we get
$$I_1+I_2=x\sqrt{x^3+1} \bigg|_{0}^{2}=6$$
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