Question

evaluate the integral using an appropriate

$\int \frac { x ^ { 6 } + 1 } { \sqrt { x ^ { 7 } + 7 x } } d x = \square + c$

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Explanation:

Step1: Let \(u = x^{7}+7x\)

Differentiate \(u\) with respect to \(x\). Using the power rule \((x^n)^\prime=nx^{n - 1}\), we have \(du=(7x^{6}+7)dx = 7(x^{6}+1)dx\), so \((x^{6}+1)dx=\frac{1}{7}du\)

Step2: Substitute into the integral

The integral \(\int\frac{x^{6}+1}{\sqrt{x^{7}+7x}}dx\) becomes \(\int\frac{1}{\sqrt{u}}\cdot\frac{1}{7}du\). Since \(\int\frac{1}{\sqrt{u}}du=\int u^{-\frac{1}{2}}du\)
Using the power rule for integration \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C\) (\(n
eq - 1\)), for \(n=-\frac{1}{2}\), \(\int u^{-\frac{1}{2}}du=\frac{u^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+C = 2u^{\frac{1}{2}}+C\)

Step3: Back - substitute \(u\)

\(\frac{1}{7}\int u^{-\frac{1}{2}}du=\frac{1}{7}\times2u^{\frac{1}{2}}+C\)
Substituting \(u = x^{7}+7x\) back in, we get \(\frac{2}{7}\sqrt{x^{7}+7x}+C\)

Answer:

\(\frac{2}{7}\sqrt{x^{7}+7x}\)