Question

Please help with the questions in the image

Answer 1

First integral:

Use the rational exponent to represent roots. You have

$\displaystyle \int\sqrt[8]{x^9}\;dx = \int x^{\frac{9}{8}}\;dx$

And from here you can use the rule

$\displaystyle \int x^n\;dx=\dfrac{x^{n+1}}{n+1}+C$

to derive

$\displaystyle \int\sqrt[8]{x^9}\;dx = \dfrac{x^{\frac{17}{8}}}{\frac{17}{8}}=\dfrac{8x^{\frac{17}{8}}}{17}$

Second integral:

Simply split the fraction:

$\dfrac{3+\sqrt{x}+x}{x}=\dfrac{3}{x}+\dfrac{\sqrt{x}}{x}+\dfrac{x}{x}=\dfrac{3}{x}+\dfrac{1}{\sqrt{x}}+1$

So, the integral of the sum becomes the sum of three immediate integrals:

$\displaystyle \int \dfrac{3}{x}\;dx = 3\log(|x|)+C$

$\displaystyle \int \dfrac{1}{\sqrt{x}}\;dx = \int x^{-\frac{1}{2}}\;dx = 2\sqrt{x}+C$

$\displaystyle \int 1\;dx = x+C$

So, the answer is the sum of the three pieces:

$3\log(|x|) + 2\sqrt{x} + x+C$

Third integral:

Again, you can split the integral of the sum in the sum of the integrals. The antiderivative of the cosine is the sine, because $\sin'(x)=\cos(x)$. So, you have

$\displaystyle \int \left( \cos(x)+\dfrac{1}{7}x\right)\;dx = \int \cos(x)\;dx + \dfrac{1}{7}\int x\;dx = \sin(x)+\frac{1}{14}x^2+C$