Finding the antiderivative of $f(x)=\sqrt[3]{x^2+2x}$

Question

$$f(x)=\sqrt[3]{x^2+2x}$$. Let $g(x)$ be an antiderivative of $f(x)$. If $g(5)=7$, then what is the value of $g(1)$?
I tried doing integrating by parts repeatedly, but no success. Wolfram Alpha also gives something in a different type of function which I don't know.
Please help?

Answer 1

$$g(x)=\int\sqrt[3]{x^2+2x}\ dx\quad\mbox{and}\quad g(5)=7$$ So now $$\int_1^5\sqrt[3]{x^2+2x}\ dx=g(5)-g(1)=7-g(1)$$ Which implies that $$g(1)=7-\int_1^5\sqrt[3]{x^2+2x}\ dx$$ Note that this integral does not have a solution in terms of elementary mathematical functions.

Answer 2

If you are confident with a numerical solution , here is it :

Because of

$$g(5)-g(1)=\int_1^5 (x^2+2x)^\frac{1}{3}dx=9.729162187801335050406060297$$

we get

$$g(1)=g(5)-9.729162187801335050406060297=-2.729162187801335050406060298$$

Answer 3

Which is the exact phrasing of the question? I mean, maybe you are not asked to compute but just to express the value in terms of $g$, i.e.: $$g(1)=\int_5^1 \sqrt[3]{x^2+2 x} \, dx +7$$ because, at the end of the day, the definition of $g$ is simply: $$g(x)=\int_5^x \sqrt[3]{t^2+2 t} \, dt +7$$