Integration by Substitution - choosing a u

Identify the u and the du:

• u=10x, du = 10dx

• u = 5x² + 1, du = 10xdx

• u = (5x² +1)², du = 10xdx

• u = (5x² +1)², du = 2(5x² + 1) 10xdx

Identify the u and the du.

• u = (3 - 4x²)³, du = 3(3-4x²)² dx

• u=4x², du = 8xdx

• u = -8x, du = -8dx

• u=3-4x², du = -8xdx

Identify the u and the du.

• u = -x², du = -2x dx

• u = 1 - x², du = -2xdx

• u = $\sqrt{1-x^2}$ , du = $-\frac{x}{\sqrt{1-x^2}}$ dx

• u = -2x, du = -2dx

Identify the u and the du.

• u = 3x², du = -6x dx

• u = x³ + 1, du = 3x² dx

• $u=\sqrt{x^{3^{ }}+1}$ , $du=\frac{3x^2}{2\sqrt{x^3+1}}$

• $u=\sqrt{x^3+1}$ , $du\ =\ 3x^2dx$

Identify the u and the du.

• $u=-\frac{2}{x^3},\ du\ =\ -\frac{2}{3x^2}dx$

• $u=\left(4+\frac{1}{x^2}\right),\ du\ =\ -\frac{2}{x^3}dx$

• $u=\frac{1}{x^2},\ du\ =\ -\frac{2}{x^3}dx$

• $u=\left(4+\frac{1}{x^2}\right)^5,\ du\ =\ 5\left(4+\frac{1}{x^2}\right)^4\times-\frac{2}{x^3}dx$

Identify the u and the du.

• u = 1+2x, du = 2dx

• u = 2, du = 0dx

• u = 2x, du = 2dx

• $u=\frac{1}{\left(1+2x\right)^2},\ du\ =\ 2dx$

Identify the u and the du.

• $u=1+\sqrt{x},\ du\ =\ \frac{1}{2\sqrt{x}}dx$

• $u=\sqrt{x},\ du\ =\ \frac{1}{2\sqrt{x}}dx$

• $u=\left(1+\sqrt{x}\right)^3,\ du\ =\ \frac{1}{2\sqrt{x}}dx$

• $u=\frac{1}{2\sqrt{x}},\ du\ =\ \sqrt{x}dx$

Identify the u, and the du.

• $u=4-\sqrt{x},\ du\ =\ -\frac{1}{2\sqrt{x}}dx$

• $u=\sqrt{x},\ du\ =\ \frac{1}{2\sqrt{x}}dx$

• $u=\frac{-1}{2\sqrt{x}},\ du\ =\ \left(4-\sqrt{x}\right)^2dx$

• $u=\left(4-\sqrt{x}\right)^2,\ du\ =\ -\frac{1}{2\sqrt{x}}dx$

Identify the u and the du.

• u = x, du = dx

• u = x -1, du = dx

• u = (x - 1)⁴, du = dx

• u = (x - 1)⁴, du = 4(x - 1)³dx

Identify the u and the du.

• u = -2x, du = -2 dx

• u = x, du = dx

• u = (7 - 2x)², du = 2(7 - 2x) dx

• u = 7 - 2x, du = -2 dx