ĐGTD ĐH Bách khoa - Tư duy Toán học - Nguyên hàm - Định nghĩa và tính chất

$\int f^{\text{'}}\left(x\right)dx=f\left(x\right)+C$

$\int f\left(x\right)dx=f\mathit{\text{'}}\left(x\right)+C$

$\int f^{\text{'}}\left(x\right)dx=f^{\text{'}\text{'}}\left(x\right)+C$

$\int f\left(x\right)dx=f^{\text{'}\text{'}}\left(x\right)+C$

$f^{\text{'}}\left(x\right)=F\left(x\right)$

$\int f\left(x\right)dx=F\left(x\right)+C$

$\int F\left(x\right)dx=f\left(x\right)+C$

$f^{\text{'}}\left(x\right)=F^{\text{'}}\left(x\right)$

y=sinx+1

y=cosx

y=cotx

y=-cosx

$\int a^{x}dx=\frac{a^x}{\ln a}+C(0<a\ne 1)$

$\int a^{x}dx=a^x+C(0<a\ne 1)$

$\int a^{x}dx=a^x\ln a+C(0<a\ne 1)$

$\int a^{x}dx=a^x\ln a(0<a\ne 1)$

$\int f\left(x\right)dx=2x+C$

$\int f\left(x\right)dx=x^2+4x+C$

$\int \left(x^2+4\right)dx=\frac{x^3}{3}+4x+C$

$\int f\left(x\right)dx=x^3+4x+C$

$y=12x^3$

$y=\frac{3x^5}{5}-1$

$y=\frac{3x^5+1}{5}$

$y=\frac{3}{5}{x}^5-\frac{3}{5}$

$\int \frac{1}{{\cos }^{2}x}dx=\tan x+C$

$\int \frac{1}{{\sin }^{2}x}dx=\cot x+C$

$\int \frac{1}{{\sin }^{2}x}dx=-\cot x+C$

$\int \frac{1}{{\cos }^{2}x}dx=\frac{\sin x}{\cos x}+C$

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

$\int dx=x+C$

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

$\int \frac{1}{x}dx=\ln \left|x\right|+C$

$\int \left(\frac{1}{{\sin }^{2}x}+\frac{1}{{\cos }^{2}x}\right)dx=\tan x-\cot x+C$

$\int \left(\frac{1}{{\sin }^{2}x}+\frac{1}{{\cos }^{2}x}\right)dx=\cot x-\tan x+C$

$\int \left(\frac{1}{{\sin }^{2}x}+\frac{1}{{\cos }^{2}x}\right)dx=\cot x+\tan x+C$

$\int \left(\frac{1}{{\sin }^{2}x}+\frac{1}{{\cos }^{2}x}\right)dx=-\cot x-\tan x+C$

$\int f\left(x\right)dx=e^{x-2}+C$

$\int f\left(x\right)dx=e^x+2x+C$

$\int f\left(x\right)dx=e^x+C$

$\int f\left(x\right)dx=e^x-2x+C$

$x+\frac{1}{x-2}+C$

$\frac{x^2}{2}+\ln \left|x-2\right|+C$

$x^2+\ln \left|x-2\right|+C$

$1+\frac{1}{{\left(x-2\right)}^2}+C$

$\frac{\ln \left(x-2\right)+\ln x}{2}+C.$

$\frac{\ln x-\ln \left(x-2\right)}{2}+C.$

$\frac{\ln \left(x-2\right)-\ln x}{2}+C.$

$\ln \left(x-2\right)-\ln x+C.$

$F\left(x\right)=\frac{1}{\sqrt{3}}-\cot x$

$F\left(x\right)=\sqrt{3}-\cot x$

$F\left(x\right)=\frac{\sqrt{3}}{2}-\cot x$

$F\left(x\right)=-\cot x+C$

$\int \frac{1}{x+2}dx=\ln \left(x+2\right)+C$

$y=\ln \left(3\left|x+2\right|\right)$ là một nguyên hàm của f(x)

$y=\ln \left|x+2\right|+\mathrm{C}$là họ nguyên hàm của f(x)

$y=\ln \left|x+2\right|\mathrm{là}\mathrm{một}\mathrm{nguyên}\mathrm{hàm}\mathrm{của}\mathrm{f}\left(\mathrm{x}\right)$

$x^2\left(1+\frac{3}{4}{x}^2\right)+C$

$\frac{x^2}{2}\left(2x+x^3\right)+C$

$x^2\left(2+6x\right)+C$

$x^2+\frac{3}{4}{x}^4$

$\int f\left(x\right)dx=\frac{x^3}{3}-\frac{2}{x}+C.$

$\int f\left(x\right)dx=\frac{x^3}{3}-\frac{1}{x}+C.$

$\int f\left(x\right)dx=\frac{x^3}{3}+\frac{2}{x}+C.$

$\int f\left(x\right)dx=\frac{x^3}{3}+\frac{1}{x}+C.$

$F\left(x\right)=-\frac{1}{2018}{e}^{-2018x+2017}+\frac{1}{2018e}$

$F\left(x\right)=-\frac{1}{2018}{e}^{-2018x+2017}+e+\frac{1}{2018e}$

$F\left(x\right)=-2018e^{-2018x+2017}+e+\frac{2018}{e}$

$F\left(x\right)=-2018e^{-2018x+2017}+\frac{1}{2018e}$

3

4

8

-1

$-\frac{2}{3}$

$-\frac{1}{2}$

-1

$-\frac{3}{4}$

P = -4

P = 1

P = -5

P = -3

$-4x^2+3x+C.$

$-4x^2+2x+C.$

$4x^2+2x+C.$

$-4x^2+x+C.$

$\frac{2}{3}\ln \left|2x+1\right|+\frac{5}{3}\ln \left|x-1\right|+C$

$-\frac{2}{3}\ln \left|2x+1\right|+\frac{5}{3}\ln \left|x-1\right|+C$

$\frac{2}{3}\ln \left|2x+1\right|-\frac{5}{3}\ln \left|x-1\right|+C$

$-\frac{1}{3}\ln \left|2x+1\right|+\frac{5}{3}\ln \left|x-1\right|+C$

${\mathrm{e}}^{-1}$

${\mathrm{e}}^2$

e

${\mathrm{e}}^{-2}$

$\sqrt{15}$

$\sqrt{23}$

$\sqrt{24}$

$\sqrt{26}$