Antiderivative and indefinite integral

The function $F(x)$ is called the antiderivative of the function $f(x)$ defined on some set $X$ if $F^{\prime }(x)=f(x)$ for all $x\in X.$ If $F(x-)$ is an antiderivative of the function $f(x),$ then $\mathrm{\Phi }(x)$ is also an antiderivative of the same function if and only if $\mathrm{\Phi }(x)=F(x)+C,$ 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. ${(\int f(x)dx)}^{\prime }=f(x).$

2. $\int f^{\prime }(x)dx=f(x)+C.$

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

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. $\int dxx=\ln |x|+C$

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

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

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

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

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

9. $\int \frac{dx}{si{n}^{2}x}=-ctgx+C$

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

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

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. $\int shxdx=chx+C$

15. $\int chxdx=shx+C$

16. $\int \frac{dx}{c{h}^{2}x}=thx+C$

17. $\int \frac{dx}{s{h}^{2}x}=-cthx+C$

Examples.

Find the antiderivatives of the following functions:

1. $2x^7.$

Solution.

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

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

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

Answer: $0.25x^8+c.$

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

Solution.

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

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

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

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

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

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

Solution.

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

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

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

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

Solution.

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

$$(tg4x{)}^{\prime }=\frac{1}{{\cos }^{2}4x}(4x{)}^{\prime }=\frac{4}{{\cos }^{2}4x}\Rightarrow (\frac{1}{4}tg4x{)}^{\prime }=\frac{1}{{\cos }^{2}4x}.$$

Thus, $$F(x)=\frac{1}{4}tg4x,$$ 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:

5.$\int (3x^2+2x+\frac{1}{x})dx.$

Solution.

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

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

6.$\int \sqrt{mx}dx.$

Solution.

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

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

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

Solution.

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

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

8.$\int 2^x{e}^{x}dx.$

Solution.

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

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

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

Solution.

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

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

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

Solution.

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

Answer: $0,5x-0,5\sin x+c.$

11.$\int \frac{dx}{\sqrt{x^2-7}}.$

Solution.

$$\int \frac{dx}{\sqrt{x^2-7}}=\ln |x+\sqrt{x^2-7}|+c.$$

Answer: $\ln |x+\sqrt{x^2-7}|+c.$

Tags: Antiderivative, Properties of the indefinite integral, Table of integrals, calculus, indefinite integral, integral, mathematical analysis