ĐGTD ĐH Bách khoa - Tư duy Toán học - Nguyên hàm (đổi biến)

dt=u'(x)dx

dx=u'(t)dt

$dt=\frac{1}{u\left(x\right)}dx$

$dx=\frac{1}{u\left(t\right)}dt$

dx=u'(t)dt

dt=u'(x)dx

dx=u(t)dt

dt=dx

$\int f\left(3x\right)\text{d}x=2x\ln \left(9x-1\right)+C.$

$\int f\left(3x\right)\text{d}x=6x\ln \left(3x-1\right)+C.$

$\int f\left(3x\right)\text{d}x=6x\ln \left(9x-1\right)+C.$

$\int f\left(3x\right)\text{d}x=3x\ln \left(9x-1\right)+C.$

$xf\left(x^2\right)dx=f\left(t\right)dt$

$xf\left(x^2\right)dx=\frac{1}{2}f\left(t\right)dt$

$xf\left(x^2\right)dx=2f\left(t\right)dt$

$xf\left(x^2\right)dx=f^2\left(t\right)dt$

$f\left(x\right)dx=-tdt$

$f\left(x\right)dx=2tdt$

$f\left(x\right)dx=-2t^{2}dt$

$f\left(x\right)dx=2t^{2}dt$

$I=\frac{4}{3}\int \left(2u^2+1\right)du$

$I=\frac{4}{3}\int \left(-u^2+1\right)du$

$I=\frac{4}{3}\int \left(u^2-1\right)du$

$I=\frac{4}{3}\int \left(\mathit{2}{u}^2-1\right)du$

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

$\int f\left(x\right)dx=\frac{1}{6}\cos 3x+\frac{1}{2}\sin x+C$

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

$\int f\left(x\right)dx=\frac{1}{6}\cos 3x-\frac{1}{2}\sin x+C$

$I=t-\frac{t^2}{2}+C$

$I=\frac{t^2}{2}-t+C$

$I=\frac{t^2}{2}-\frac{t^2}{3}+C$

$I=-\frac{t^2}{2}+\frac{t^2}{3}+C$

$x=1-\sqrt{3}$

x = 1

x = -1

x = 0

$\int f\left(x\right)\text{d}x=4\int {\cos }^{2}t\text{d}t.$

$\int f\left(x\right)\text{d}x=8\int {\cos }^{2}t\text{d}t.$

$\int f\left(x\right)\text{\hspace{0.17em}d}x=\int \left(1+\cos 2t\right)\text{\hspace{0.17em}\hspace{0.17em}d}t.$

$\int f\left(x\right)\text{\hspace{0.17em}d}x=2t-\sin 2t+C.$

$\int f(5x+2)dx=5F\left(x\right)+2+C$

$\int f(5x+2)dx=F(5x+2)+C$

$\int f(5x+2)dx=\frac{1}{5}F(5x+2)+C$

$\int f(5x+2)dx=5F(5x+2)+C$

$\ln \left|\cos x\right|+C$

$\ln \left|\sin x\right|+C$

$\sin x+C$

tan x + C

$I=\frac{1}{5}{\left(x^3+1\right)}^2\sqrt{x^3+1}-\frac{1}{3}\left(x^3+1\right)\sqrt{x^3+1}+C$

$I=\frac{2}{5}{\left(x^3+1\right)}^2\sqrt{x^3+1}-\frac{2}{3}\left(x^3+1\right)\sqrt{x^3+1}+C$

$I=\frac{2}{5}{\left(x^3+1\right)}^2\sqrt{x^3+1}+C$

$I=\frac{2}{5}{\left(x^3+1\right)}^2\sqrt{x^3+1}+\left(x^3+1\right)\sqrt{x^3+1}+C$

$F\left(x\right)=-2\sqrt{1-\ln x}+\frac{2}{3}\left(1-\ln x\right)\sqrt{1-\ln x}+3$

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

$F\left(x\right)=-2\sqrt{1-\ln x}-\frac{2}{3}\left(1-\ln x\right)\sqrt{1-\ln x}+3$

$F\left(x\right)=2\sqrt{1-\ln x}-\frac{2}{3}\left(1-\ln x\right)\sqrt{1-\ln x}+3$

m = 2

m = -2

m = 1

m = -1

6

4

8

5

-2

2

-1

1

$\int f\left(x\right)d\text{x}=\frac{1}{3}\sqrt{3{\text{x}}^2+2}+C$

$\int f\left(x\right)d\text{x}=-\frac{1}{3}\sqrt{3{\text{x}}^2+2}+C$

$\int f\left(x\right)d\text{x}=\frac{1}{6}\sqrt{3{\text{x}}^2+2}+C$

$\int f\left(x\right)d\text{x}=\frac{2}{3}\sqrt{3{\text{x}}^2+2}+C$

$I=-\int {\cos }^{2}t\text{\hspace{0.17em}\hspace{0.17em}d}t.$

$I=\int {\sin }^{2}t\text{\hspace{0.17em}\hspace{0.17em}d}t.$

$I=\int {\cos }^{2}t\text{\hspace{0.17em}\hspace{0.17em}d}t.$

$I=\frac{1}{2}\int \left(1+\cos 2t\right)\text{d}t.$

$\frac{{\pi }^2}{2}+\ln \frac{\pi }{2}+1$

$\frac{{\pi }^2}{4}-\ln \frac{\pi }{2}+1.$

$\frac{{\pi }^2}{8}.$

$\frac{{\pi }^2}{8}+\ln \frac{\pi }{2}+1.$

3

4

1

2