Properties of Indefinite Integrals

Example 1. Find $$${F}{\left({x}\right)}=\int{f{{\left({x}\right)}}}{d}{x}$$$ if $$${f{{\left({x}\right)}}}={{e}}^{{x}}+{{x}}^{{-\frac{{2}}{{3}}}}-\frac{{5}}{{{1}+{{x}}^{{2}}}}$$$.

Using table of antiderivatives and Properties 1 and 2 we have that $$${F}{\left({x}\right)}=\int{\left({{e}}^{{x}}+{{x}}^{{-\frac{{2}}{{3}}}}-\frac{{5}}{{{1}+{{x}}^{{2}}}}\right)}{d}{x}=\int{{e}}^{{x}}{d}{x}+\int{{x}}^{{-\frac{{2}}{{3}}}}{d}{x}-{5}\int\frac{{1}}{{{1}+{{x}}^{{2}}}}{d}{x}=$$$

$$$={{e}}^{{x}}+\frac{{{{x}}^{{-\frac{{2}}{{3}}+{1}}}}}{{-\frac{{2}}{{3}}+{1}}}+{5}{\operatorname{arctan}{{\left({x}\right)}}}+{C}$$$.

So $$${F}{\left({x}\right)}={{e}}^{{x}}+{3}{{x}}^{{\frac{{1}}{{3}}}}+{5}{\operatorname{arctan}{{\left({x}\right)}}}+{C}$$$.

Example 2. Find $$${f{{\left({x}\right)}}}$$$ if $$${f{'}}{\left({x}\right)}={\cos{{\left({x}\right)}}}+{\sin{{\left({x}\right)}}}-{5}{{x}}^{{2}}$$$ and $$${f{{\left({0}\right)}}}={2}$$$.

According to Property 4 $$${f{{\left({x}\right)}}}=\int{f{'}}{\left({x}\right)}{d}{x}+{C}$$$.

So $$${f{{\left({x}\right)}}}=\int{\left({\cos{{\left({x}\right)}}}+{\sin{{\left({x}\right)}}}-{5}{{x}}^{{2}}\right)}{d}{x}=\int{\cos{{\left({x}\right)}}}{d}{x}+\int{\sin{{\left({x}\right)}}}{d}{x}-{5}\int{{x}}^{{2}}{d}{x}=$$$

$$$={\sin{{\left({x}\right)}}}-{\cos{{\left({x}\right)}}}-\frac{{5}}{{3}}{{x}}^{{3}}+{C}$$$.

Since $$${f{{\left({0}\right)}}}={2}$$$ then $$${2}={\sin{{\left({0}\right)}}}-{\cos{{\left({0}\right)}}}-\frac{{5}}{{3}}{{\left({0}\right)}}^{{3}}+{C}$$$ or $$${C}={3}$$$.

Therefore $$${f{{\left({x}\right)}}}={\sin{{\left({x}\right)}}}-{\cos{{\left({x}\right)}}}-\frac{{5}}{{3}}{{x}}^{{3}}+{3}$$$.

Example 3. Find $$$\int{{\left({2}{{x}}^{{2}}+{1}\right)}}^{{3}}{d}{x}$$$.

Using the formula of Binom of Newton we can rewrite integrand as follows:

$$$\int{{\left({2}{{x}}^{{2}}+{1}\right)}}^{{3}}{d}{x}=\int{\left({{\left({2}{{x}}^{{2}}\right)}}^{{3}}+{3}{{\left({2}{{x}}^{{2}}\right)}}^{{2}}\cdot{1}+{3}\cdot{2}{{x}}^{{2}}\cdot{{1}}^{{2}}+{{1}}^{{3}}\right)}{d}{x}=\int{\left({8}{{x}}^{{6}}+{12}{{x}}^{{4}}+{6}{{x}}^{{2}}+{1}\right)}{d}{x}=$$$

$$$=\frac{{8}}{{7}}{{x}}^{{7}}+\frac{{12}}{{5}}{{x}}^{{5}}+{2}{{x}}^{{3}}+{x}+{C}$$$.

Example 4. Find $$$\int{{\left({1}+\sqrt{{{x}}}\right)}}^{{4}}{d}{x}$$$.

Using formula of Binom of Newton we have that

$$$\int{{\left({1}+\sqrt{{{x}}}\right)}}^{{4}}{d}{x}=\int{\left({1}+{4}\sqrt{{{x}}}+{6}{x}+{4}{x}\sqrt{{{x}}}+{{x}}^{{2}}\right)}{d}{x}=$$$

$$$=\int{1}{d}{x}+{4}\int{{x}}^{{\frac{{1}}{{2}}}}{d}{x}+{6}\int{x}{d}{x}+{4}\int{{x}}^{{\frac{{3}}{{2}}}}{d}{x}+\int{{x}}^{{2}}{d}{x}={x}+\frac{{8}}{{3}}{{x}}^{{\frac{{3}}{{2}}}}+{3}{{x}}^{{2}}+\frac{{8}}{{5}}{{x}}^{{\frac{{5}}{{2}}}}+\frac{{1}}{{3}}{{x}}^{{3}}+{C}$$$.

Example 5. Find $$$\int\frac{{{\left({x}+{1}\right)}{\left({{x}}^{{2}}-{3}\right)}}}{{{3}{{x}}^{{2}}}}{d}{x}$$$.

We have that $$$\int\frac{{{{x}}^{{3}}+{{x}}^{{2}}-{3}{x}-{3}}}{{{3}{{x}}^{{2}}}}{d}{x}=\int{\left(\frac{{1}}{{3}}{x}+\frac{{1}}{{3}}-\frac{{1}}{{x}}-\frac{{1}}{{{x}}^{{2}}}\right)}{d}{x}=\frac{{1}}{{3}}\int{x}{d}{x}+\frac{{1}}{{3}}\int{1}\cdot{d}{x}-\int\frac{{1}}{{x}}{d}{x}-\int\frac{{1}}{{{x}}^{{2}}}{d}{x}=$$$

$$$=\frac{{1}}{{6}}{{x}}^{{2}}+\frac{{1}}{{3}}{x}-{\ln}{\left|{x}\right|}+\frac{{1}}{{x}}+{C}$$$.

Example 6. Find $$$\int\frac{{{\left({x}-\sqrt{{{x}}}\right)}{\left({1}+\sqrt{{{x}}}\right)}}}{{{\sqrt[{{3}}]{{x}}}}}{d}{x}$$$.

We have that $$$\int\frac{{{\left({x}-\sqrt{{{x}}}\right)}{\left({1}+\sqrt{{{x}}}\right)}}}{{{\sqrt[{{3}}]{{x}}}}}{d}{x}=\int\frac{{{x}+{x}\sqrt{{{x}}}-\sqrt{{{x}}}-{x}}}{{{\sqrt[{{3}}]{{x}}}}}{d}{x}=\int\frac{{{x}\sqrt{{{x}}}-\sqrt{{{x}}}}}{{{\sqrt[{{3}}]{{x}}}}}{d}{x}=\int\frac{{{{x}}^{{\frac{{3}}{{2}}}}-{{x}}^{{\frac{{1}}{{2}}}}}}{{{{x}}^{{\frac{{1}}{{3}}}}}}{d}{x}=$$$

$$$=\int{\left({{x}}^{{\frac{{7}}{{6}}}}-{{x}}^{{\frac{{1}}{{6}}}}\right)}{d}{x}=\int{{x}}^{{\frac{{7}}{{6}}}}{d}{x}-\int{{x}}^{{\frac{{1}}{{6}}}}{d}{x}=\frac{{6}}{{13}}{{x}}^{{\frac{{13}}{{6}}}}-\frac{{6}}{{7}}{{x}}^{{\frac{{7}}{{6}}}}+{C}$$$.

Example 7. Find $$${f{{\left({x}\right)}}}$$$ if $$${f{''}}{\left({x}\right)}={\cos{{\left({x}\right)}}}+{{x}}^{{-\frac{{3}}{{5}}}}$$$ and $$${f{{\left({0}\right)}}}={2}$$$ and $$${f{{\left({1}\right)}}}={3}$$$.

We have that $$${f{'}}{\left({x}\right)}=\int{f{''}}{\left({x}\right)}{d}{x}$$$.

That's why $$${f{'}}{\left({x}\right)}={\sin{{\left({x}\right)}}}+\frac{{{{x}}^{{-\frac{{3}}{{5}}+{1}}}}}{{-\frac{{3}}{{5}}+{1}}}+{C}$$$ or $$${f{'}}{\left({x}\right)}={\sin{{\left({x}\right)}}}+\frac{{5}}{{2}}{{x}}^{{\frac{{2}}{{5}}}}+{C}$$$.

Now $$${f{{\left({x}\right)}}}=\int{f{'}}{\left({x}\right)}{d}{x}$$$, so

$$${f{{\left({x}\right)}}}=-{\cos{{\left({x}\right)}}}+\frac{{5}}{{2}}\frac{{{{x}}^{{\frac{{2}}{{5}}+{1}}}}}{{\frac{{2}}{{5}}+{1}}}+{C}{x}+{D}$$$ or $$${f{{\left({x}\right)}}}=-{\cos{{\left({x}\right)}}}+\frac{{25}}{{14}}{{x}}^{{\frac{{7}}{{5}}}}+{C}{x}+{D}$$$.

Note that both $$${C}$$$ and $$${D}$$$ are constants.

Since $$${f{{\left({0}\right)}}}={2}$$$ then $$${2}=-{\cos{{\left({0}\right)}}}+\frac{{25}}{{14}}{{\left({0}\right)}}^{{\frac{{7}}{{5}}}}+{C}\cdot{0}+{D}$$$ and $$${D}={3}$$$.

Now, $$${f{}}$$$ has the following form: $$${f{{\left({x}\right)}}}=-{\cos{{\left({x}\right)}}}+\frac{{25}}{{14}}{{x}}^{{\frac{{7}}{{5}}}}+{C}{x}+{3}$$$.

Since $$${f{{\left({1}\right)}}}={3}$$$ then $$${3}=-{\cos{{\left({1}\right)}}}+\frac{{25}}{{14}}{{\left({1}\right)}}^{{\frac{{7}}{{5}}}}+{C}\cdot{1}+{3}$$$ or $$${C}={\cos{{\left({1}\right)}}}-\frac{{25}}{{14}}$$$.

Therefore $$${f{{\left({x}\right)}}}=-{\cos{{\left({x}\right)}}}+\frac{{25}}{{14}}{{x}}^{{\frac{{7}}{{5}}}}+{\left({\cos{{\left({1}\right)}}}-\frac{{25}}{{14}}\right)}{x}+{3}$$$.

Example 8. A particle moves in a straight line and has acceleration given by $$${a}{\left({t}\right)}={2}{t}+{3}$$$. It is initial velocity is $$${v}{\left({0}\right)}=-{4}\frac{{m}}{{s}}$$$ and its initial dispalcement is $$${s}{\left({0}\right)}={10}\ {c}{m}$$$. Find its position function.

Since $$${v}'{\left({t}\right)}={a}{\left({t}\right)}={2}{t}+{3}$$$ then integration gives $$${v}{\left({t}\right)}={2}\frac{{{t}}^{{2}}}{{2}}+{3}{t}+{C}={{t}}^{{2}}+{3}{t}+{C}$$$.

We have $$${v}{\left({0}\right)}=-{4}$$$, so $$$-{4}={{0}}^{{2}}+{3}\cdot{0}+{C}$$$ or $$${C}=-{4}$$$, therefore $$${v}{\left({t}\right)}={{t}}^{{2}}+{3}{t}-{4}$$$.

Since $$${s}'{\left({t}\right)}={v}{\left({t}\right)}={{t}}^{{2}}+{3}{t}-{4}$$$ then integration gives $$${v}{\left({t}\right)}=\frac{{{{t}}^{{3}}}}{{3}}+{3}\frac{{{{t}}^{{2}}}}{{2}}-{4}{t}+{D}$$$.

We have $$${s}{\left({0}\right)}={10}$$$, so $$${10}=\frac{{1}}{{3}}\cdot{{0}}^{{3}}+\frac{{3}}{{2}}\cdot{{0}}^{{2}}-{4}\cdot{0}+{D}$$$ or $$${D}={10}$$$.

Therefore position function is $$${s}{\left({t}\right)}=\frac{{1}}{{3}}{{t}}^{{3}}+\frac{{3}}{{2}}{{t}}^{{2}}-{4}{t}+{10}$$$.

Example 9. A ball is thrown upward with a speed of $$${64}\ \frac{{f{{t}}}}{{s}}$$$ from the edge of a cliff 464 ft above the ground. Find its height above the ground $$${t}$$$ seconds later. When does it reach its maximum height? When does it hit the ground?

An object near the surface of the earth is subject to a gravitational force that produces a downward acceleration denoted by $$${g{}}$$$.

For motion close to the earth we may assume that $$${g{}}$$$ is constant, its value being about $$${9.8}\frac{{m}}{{{s}}^{{2}}}$$$ (or $$${32}\frac{{f{{t}}}}{{{s}}^{{2}}}$$$).

The motion is vertical and we choose the positive direction to be upward. At time $$${t}$$$ the distance above the ground is $$${s}{\left({t}\right)}$$$ and the velocity $$${v}{\left({t}\right)}$$$ is decreasing.
Therefore, the acceleration $$${a}{\left({t}\right)}$$$ must be negative and we have $$${a}{\left({t}\right)}=\frac{{{d}{v}}}{{{d}{t}}}=-{32}$$$.

Taking integral, we have $$${v}{\left({t}\right)}=-{32}{t}+{C}$$$.

From the fact that $$${v}{\left({0}\right)}={64}$$$ we have that $$${64}=-{32}\cdot{0}+{C}$$$ or $$${C}={64}$$$.

Therefore, $$${v}{\left({t}\right)}=-{32}{t}+{64}$$$.

The maximum height will be reached when $$${v}{\left({t}\right)}={0}$$$ or $$${t}=\frac{{64}}{{32}}={2}\ {s}$$$.

Since $$${s}'{\left({t}\right)}={v}{\left({t}\right)}$$$, we integrate again and obtain $$${s}{\left({t}\right)}=-{16}{{t}}^{{2}}+{64}{t}+{D}$$$.

Using the fact that $$${s}{\left({0}\right)}={464}$$$ we have that $$${464}=-{16}\cdot{{0}}^{{2}}+{64}\cdot{0}+{D}$$$ or $$${D}={464}$$$.

Therefore, $$${s}{\left({t}\right)}=-{16}{{t}}^{{2}}+{64}{t}+{464}$$$.

The expression for $$${s}{\left({t}\right)}$$$ is valid until the ball hits the ground. This happens when $$${s}{\left({t}\right)}={0}$$$, that is, when $$$-{16}{{t}}^{{2}}+{64}{t}+{464}={0}$$$ or $$${{t}}^{{2}}-{4}{t}-{29}={0}$$$.

Using the quadratic formula to solve this equation, we get $$${t}={2}\pm\sqrt{{{33}}}$$$.

We reject the solution with the minus sign since it gives a negative value for time $$${t}$$$.
Therefore, the ball hits the ground after $$${2}+\sqrt{{{33}}}\approx{7.7}\ {s}.$$$