Antiderivative and indefinite integral

The function $F(x)$ is called the antiderivative of the function $f(x)$ defined on some set $X$ if $F'(x)=f(x)$ for all $x\in X.$ If $F(x -)$ is an antiderivative of the function $f(x),$ then $\Phi(x)$ is also an antiderivative of the same function if and only if $\Phi(x)=F(x)+C,$ where $C$ is some constant. The set of all antiderivatives of the function $f(x)$ is called the indefinite integral of this function and is denoted by the symbol.

$$\int f(x)\,dx.$$ Thus, by definition $$\int f(x)\,dx=F(x)+C,$$ where $F(x)$ is one of the antiderivatives of the function $f(x)$ and the constant $C$ takes real values.

Properties of the indefinite integral.

1. $\left(\int f(x)\,dx\right)'=f(x).$

2. $\int f'(x)dx=f(x)+C.$

3. $\int af(x)dx=a\int f(x) dx.\,\,\,\,\,\,a\neq 0.$

4. $\int (f_1(x)+f_2(x))dx=\int f_1(x)\,dx+\int f_2(x)\, dx.$

Table of basic indefinite integrals.

1. $\int dx=x+C$

2. $\int x^{\alpha}dx=\frac{x^{\alpha+1}}{\alpha+1}+C$

3. $\int {dx}{x}=\ln |x|+C$

4. $\int a^x dx=\frac{a^x}{\ln a}+C$

5. $\int e^x dx=e^x+C$

6. $\int \sin x dx=-\cos x+C$

7. $\int \cos x dx=\sin x+C$

8. $\int \frac{dx}{\cos^2 x}=tg x+C$

9. $\int \frac{dx}{sin^2 x}=-ctg x+C$

10. $\int \frac{dx}{\sqrt{a^2-x^2}}=\arcsin\frac{x}{a}+C$

11. $\int \frac{dx}{\sqrt{x^2 \pm a^2}}=\ln\left|x+\sqrt{x^2\pm a^2}\right|+C$

12. $\int \frac{dx}{x^2+a^2}=\frac{1}{a}arctg\frac{x}{a}+C$

13. $\int \frac{dx}{x^2 -a^2}=\frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right|+C$

14. $\int sh x dx = ch x+C$

15. $\int ch x dx = sh x+C$

16. $\int \frac{dx}{ch^2 x} = th x+C$

17. $\int \frac{dx}{sh^2 x} = -cth x+C$

Examples.

Find the antiderivatives of the following functions:

6.1. $2x^7.$

Solution.

From the definition of antiderivative, we need to find a function $F(x)$ such that $F'(x)=2x^7.$

$$(x^8)'=8x^7\Rightarrow (\frac{1}{4}x^8)'=2x^7.$$

Thus, $F(x)=0.25x^8,$ and all antiderivatives of the given function have the form $0.25x^8+c.$

Answer: $0.25x^8+c.$

6.4. $\frac{x^3+5x^2-1}{x}.$

Solution.

From the definition of antiderivative, we need to find a function $F(x)$ such that $F'(x)=\frac{x^3+5x^2-1}{x}=x^2+5x-\frac{1}{x}.$

$$(x^3)'=3x^2\Rightarrow (\frac{1}{3}x^3)'=x^2;$$

$$(x^2)'=2x\Rightarrow (\frac{5}{2}x^2)'=5x;$$

$$(\ln |x|)'=\frac{1}{x}.$$

From here, we find $$F(x)=\frac{1}{3} x^3+\frac{5}{2}x^2-\ln |x|,$$ and all antiderivatives of the given function have the form $\frac{1}{3} x^3+\frac{5}{2}x^2-\ln |x|+c.$

Answer: $\frac{1}{3} x^3+\frac{5}{2}x^2-\ln |x|+c.$

6.7. $\frac{1}{\sqrt{a+bx}}.$

Solution.

From the definition of antiderivative, we need to find a function $F(x)$ such that $F'(x)=\frac{1}{\sqrt{a+bx}}.$

$$(\sqrt{a+bx})'=\frac{1}{2\sqrt{a+bx}}(a+bx)'=\frac{b}{2\sqrt{a+bx}}\Rightarrow$$ $$\Rightarrow (\frac{2}{b}\sqrt{a+bx})'=\frac{1}{\sqrt{a+bx}}.$$

Thus, $$F(x)=\frac{2}{b}\sqrt{a+bx},$$ and all antiderivatives of the given function have the form $\frac{2}{b}\sqrt{a+bx}+c.$

Answer: $\frac{2}{b}\sqrt{a+bx}+c.$

6.10. $\frac{1}{\cos^2{4x}}.$

Solution.

From the definition of antiderivative, we need to find a function $F(x)$ such that $F'(x)=\frac{1}{\cos^2{4x}}.$

$$(tg 4x)'=\frac{1}{\cos^2{4x}}(4x)'=\frac{4}{\cos^2{4x}}\Rightarrow (\frac{1}{4}tg 4x)'=\frac{1}{\cos^2 4x}.$$

Thus, $$F(x)=\frac{1}{4 }tg 4x,$$ and all antiderivatives of the given function have the form $0.25 \tan 4x+c$.

Answer: $0.25 \tan 4x+c$.

Using the table of basic integrals, find the following integrals:

6.15.$\int\left(3x^2+2x+\frac{1}{x}\right)\, dx.$

Solution.

$$\int\left(3x^2+2x+\frac{1}{x}\right)\, dx=3\int x^2 dx+2\int xdx+\int\frac{1}{x}dx=$$ $$=3\frac{x^3}{3}+2\frac{x^2}{2}+\ln |x|+c=x^3+x^2+\ln|x|+c.$$

Answer: $x^3+x^2+\ln|x|+c.$

6.17.$\int\sqrt{mx}\,dx.$

Solution.

$$\int\sqrt{mx}\, dx=\sqrt m\int x^{\frac{1}{2}}\,dx=\sqrt m\frac{x^{1/2+1}}{1/2+1}+c=\sqrt m\frac{x^{\frac{3}{2}}}{\frac{3}{2}}+c=$$ $$=\frac{2\sqrt {mx^3}}{3}+c.$$

Answer: $\frac{2\sqrt{mx^3}}{3}+c.$

6.19.$\int\left(\frac{1}{\sqrt[3]{x^2}}-\frac{x+1}{\sqrt[4]{x^3}}\right)\,dx.$

Solution.

$$\int\left(\frac{1}{\sqrt[3]{x^2}}-\frac{x+1}{\sqrt[4]{x^3}}\right)\,dx=\int x^{-2/3}dx-\int\frac{x}{x^{3/4}}\,dx-\int\frac{1}{x^{3/4}}dx=$$ $$=\int x^{-2/3}dx-\int{x^{1/4}}\,dx-\int{x^{-3/4}}dx=$$ $$=\frac{x^{-2/3+1}}{-2/3+1}-\frac{x^{1/4+1}}{1/4+1}-\frac{x^{-3/4+1}}{-3/4+1}+c=$$ $$=3x^{1/3}-\frac{4x^{5/4}}{5}-4{x^{1/4}}+c=$$

Answer: $3\sqrt[3]x-\frac{4}{5}x\sqrt[4]{x}-4\sqrt[4]{x}+c.$

6.22.$\int 2^xe^x\, dx.$

Solution.

$$\int 2^xe^x\,dx=\int (2e)^x\,dx=\frac{(2e)^x}{\ln (2e)}+c=\frac{(2e)^x}{\ln2+1}+c$$

Answer: $\frac{(2e)^x}{\ln 2+1}+c.$

6.24.$\int(2x+3\cos x)\,dx.$

Solution.

$$\int (2x+3\cos x)\,dx=2\int x\,dx+3\int\cos x\,dx=2\frac{x^2}{2}+3\sin x+c=$$ $$=x^2+3\sin x+c$$

Answer: $x^2+3\sin x+c.$

6.28.$\int\sin^2\frac{x}{2}\,dx.$

Solution.

$$\int \sin^2\frac{x}{2}\,dx=\int \frac{1-\cos x}{2}\,dx=\frac{1}{2}\int\,dx-\frac{1}{2}\int\cos x\,dx=$$ $$=0,5 x-0,5\sin x+c.$$