Show that $\int_{-1}^{1} \frac{1-x^2-x^3-x^{30}}{\sqrt[5]{(7-5x^2-x^{30})^4}}~dx<\frac{1}{3}$

Question

By evaluating the integral or otherwise can one show by hand that $$\int_{-1}^{1} \frac{1-x^2-x^3-x^{30}}{\sqrt[5]{(7-5x^2-x^{30})^4}}~dx<\frac{1}{3}?$$

Answer 1

Let $$I=\int_{-1}^{1} \frac{1-x^2-x^3-x^{30}}{\sqrt[5]{(7-5x^2-x^{30})^4}}~dx=2\int_{0}^{1} \frac{1-x^2-x^{30}}{\sqrt[5]{(7-5x^2-x^{30})^4}}~dx =2\int_{0}^{1} \frac{(x^4-x^6-x^{34})}{\sqrt[5]{(7x^5-5x^7-x^{35})^4}}~dx.$$ In the last integral, we have multiplied the integrand up and down by $x^4$. Next, letting $7x^5-5x^7-x^{35}=t$, we get $$I=\frac{2}{35}\int_{0}^{1} \frac{dt}{t^{4/5}}=\frac{2}{7} <\frac{1}{3}.$$