Integration by substitution 2

What should be used for u in the integral?

• no u needed
• 4t²
• t³
• 1/4t²

Tags

CCSS.HSA.APR.A.1

CCSS.HSA.APR.D.6

CCSS.HSA.SSE.A.1

CCSS.HSA.SSE.A.2

CCSS.HSA.SSE.B.3

 $\int15x^4\sqrt{3x^5-2}dx$  

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

•  $3x^5\sqrt{\frac{1}{2}x^6-2x}+C$  

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

•  $\sqrt{\left(3x^5-2\right)^3}+C$  

Given  $\frac{d}{dx}\left(\frac{2}{x-3}\right)=g\left(x\right)$  , find  $\int_{ }^{ }g\left(x\right)dx$  .

•  $\frac{2}{x-3}$  

•  $\frac{2}{x+3}$  

•  $-\frac{2}{x-3}$  

•  $\frac{2}{-x-3}$  

 $\int_{ }8e^{7+2x}dx$  

•  $8\left(\frac{e^{7+^{ }2x}}{2\left(\ln e\right)}\right)$  

•  $8\left(\frac{e^{7+^{ }2x}}{\left(\ln e\right)}\right)+c$  

•  $8\left(\frac{e^{7+^{ }2x}}{2\left(\ln e\right)}\right)+c$  

•  $\left(\frac{e^{7+^{ }2x}}{2\left(\ln e\right)}\right)+c$  

 $\int_{ }\frac{4}{5x-6}dx$  

•  $4\ln\left|5x-6\right|+c$  

•  $\frac{4}{5}\ln\left|5x-6\right|+c$  

•  $\frac{4}{5}\ln\left|5x-6\right|$  

•  $\ln\left|5x-6\right|+c$  

 $\int_{ }\frac{\left(\ln3x\right)^4}{x}dx$  by using method of substitution.

•  $\frac{u^5}{5}$  

•  $\frac{\left(\ln3x\right)^5}{5}+c$  

•  $\ln3x+c$  

•  $\frac{\ln3x}{5}+c$  

Tags

CCSS.HSF.BF.B.5

 $\int_{ }\left(3\tan x+4\right)^5\sec^2x\ dx$  by using method of substitution.

•  $\left(3\tan x\ +4\right)^6+c$  

•  $\frac{\left(3\tan x\ +4\right)^6}{6}$  

•  $\frac{\left(3\tan x\ -4\right)^6}{6}+c$  

•  $\frac{\left(3\tan x\ +4\right)^6}{6}+c$  

Integrate by using substitution  $\int\ \frac{2}{x\ln x}\ dx$  

•  $\frac{2}{x}+c$  

•  $2\ln\left|u\right|+c$  

•  $2\ln\left|\ln x\right|+c$  

•  $2\ln\left|x\right|+c$  

Evaluate  $\int_1^3\ \frac{x}{1+2x^2}dx$  

•  $0.4615$  

•  $\frac{1}{4}\ln\left|16\right|$  

•  $0.1425$  

•  $\ln\left|\frac{19}{3}\right|$  

If  $\int_1^3\ f\left(x\right)\ dx\ =5,\ \int_1^{11}\ f\left(x\right)\ dx=24$  and  $\int_1^3\ g\left(x\right)\ dx=7,$  evaluate  $\int_3^{11}\ f\left(x\right)\ dx$  

•  $-19$  

•  $41$  

•  $-11$  

•  $19$