Question

1. $int_{1}^{-4}\frac{1}{sqrt{x}(4sqrt{x})}dx$ 5. $int_{1}^{5}\frac{4z^{2}-1}{2z - 1}dz$

Explanation:

Step1: Simplify the first - integral

For the integral $\int_{1}^{4}\frac{1}{\sqrt{x}(4 + \sqrt{x})}dx$, use substitution. Let $u = 4+\sqrt{x}$, then $du=\frac{1}{2\sqrt{x}}dx$. When $x = 1$, $u=4 + 1=5$; when $x = 4$, $u=4 + 2=6$. The integral becomes $2\int_{5}^{6}\frac{1}{u}du$.

Step2: Evaluate the first - integral

Using the formula $\int\frac{1}{u}du=\ln|u|+C$, we have $2\int_{5}^{6}\frac{1}{u}du=2[\ln u]_{5}^{6}=2(\ln6-\ln5)=2\ln\frac{6}{5}$.

Step3: Simplify the second - integral

For the integral $\int_{1}^{5}\frac{4z^{2}-1}{2z - 1}dz$, factor the numerator: $4z^{2}-1=(2z + 1)(2z - 1)$. Then $\frac{4z^{2}-1}{2z - 1}=2z + 1$ for $z
eq\frac{1}{2}$.

Step4: Evaluate the second - integral

Using the power - rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n
eq - 1)$, we have $\int_{1}^{5}(2z + 1)dz=[z^{2}+z]_{1}^{5}=(5^{2}+5)-(1^{2}+1)=(25 + 5)-(1 + 1)=28$.

Step5: Combine the results

Since the two integrals are separate, we have the results of the two integrals: The value of $\int_{1}^{4}\frac{1}{\sqrt{x}(4+\sqrt{x})}dx$ is $2\ln\frac{6}{5}$ and the value of $\int_{1}^{5}\frac{4z^{2}-1}{2z - 1}dz$ is $28$.

Answer:

The value of $\int_{1}^{4}\frac{1}{\sqrt{x}(4+\sqrt{x})}dx$ is $2\ln\frac{6}{5}$ and the value of $\int_{1}^{5}\frac{4z^{2}-1}{2z - 1}dz$ is $28$.