Question

1. find the indicated antiderivatives

a) (int 3 sqrt 3 { x } + 2 cos x - sec ^ { 2 } x d x)
b) (int \frac { 3 } { x ^ { 4 } } + 4 e ^ { - 4 x } + \frac { 1 } { sqrt { x } } d x)

Explanation:

Step1: Use integral rules

Use the sum - rule of integration \(\int(f(x)+g(x)+h(x))dx=\int f(x)dx+\int g(x)dx+\int h(x)dx\), power - rule \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n
eq - 1)\), \(\int\cos xdx=\sin x + C\), \(\int\sec^{2}xdx=\tan x + C\), \(\int e^{ax}dx=\frac{e^{ax}}{a}+C(a
eq0)\)

Part a)

• Integrate each term separately:
• For the term \(3\sqrt[3]{x}=3x^{\frac{1}{3}}\), using the power - rule \(\int3x^{\frac{1}{3}}dx=3\times\frac{x^{\frac{1}{3}+1}}{\frac{1}{3}+1}+C_1\). Since \(\frac{1}{3}+1=\frac{4}{3}\), then \(3\times\frac{x^{\frac{4}{3}}}{\frac{4}{3}}=\frac{9}{4}x^{\frac{4}{3}}\)
• For the term \(2\cos x\), using \(\int\cos xdx=\sin x + C\), we have \(\int2\cos xdx=2\sin x + C_2\)
• For the term \(-\sec^{2}x\), using \(\int\sec^{2}xdx=\tan x + C\), we have \(\int(-\sec^{2}x)dx=-\tan x + C_3\)

• Combine the results:

\(\int(3\sqrt[3]{x}+2\cos x-\sec^{2}x)dx=\frac{9}{4}x^{\frac{4}{3}}+2\sin x-\tan x + C\) (where \(C = C_1 + C_2+C_3\))

Part b)

• Rewrite the terms:

\(\frac{3}{x^{4}}=3x^{-4}\) and \(\frac{1}{\sqrt{x}}=x^{-\frac{1}{2}}\)

• Integrate each term separately:
• For the term \(3x^{-4}\), using the power - rule \(\int3x^{-4}dx=3\times\frac{x^{-4 + 1}}{-4+1}+C_1\). Since \(-4 + 1=-3\), then \(3\times\frac{x^{-3}}{-3}=-x^{-3}\)
• For the term \(4e^{-4x}\), using \(\int e^{ax}dx=\frac{e^{ax}}{a}+C\), we have \(\int4e^{-4x}dx=4\times\frac{e^{-4x}}{-4}+C_2=-e^{-4x}\)
• For the term \(x^{-\frac{1}{2}}\), using the power - rule \(\int x^{-\frac{1}{2}}dx=\frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+C_3\). Since \(-\frac{1}{2}+1=\frac{1}{2}\), then \(\frac{x^{\frac{1}{2}}}{\frac{1}{2}} = 2x^{\frac{1}{2}}\)

• Combine the results:

\(\int(\frac{3}{x^{4}}+4e^{-4x}+\frac{1}{\sqrt{x}})dx=-x^{-3}-e^{-4x}+2x^{\frac{1}{2}}+C\) (where \(C = C_1 + C_2+C_3\))

Answer:

a) \(\frac{9}{4}x^{\frac{4}{3}}+2\sin x-\tan x + C\)
b) \(-\frac{1}{x^{3}}-e^{-4x}+2\sqrt{x}+C\)