${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}(x^3+1{)}^{20}3x^{2}dx=$

poinits$/25$ pointe en

A$)\frac{(x^3+1{)}^{21}}{21}+C$

B$)\frac{(x^3+1{)}^{21}}{21}3x^2+C$

C$)$

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}4x^{3}sin(x^4)dx=$

A$)-cos(x^4)+C$

B$)cos(x^4)+C$

C$)\frac{cos(r^2)}{4}+C$

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}(2x+1{)}^{7}dx=$

A$)\frac{(2x+1{)}^8}{16}+C$

B$)\frac{(2x+1{)}^n}{8}+C$

C$)\frac{(2a+3{)}^n}{16}$

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{e^x}{1+e^{2x}}dx=$

A$)si{n}^{-1}(e^x)+C$

B$)co{s}^{-1}(e^x)+C$

C$)ta{n}^{-1}(e^2)+C$

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{dx}{\sqrt[\phantom{2}]{5-x^2}}=$

A$)co{s}^{-1}(\frac{x}{\sqrt[\phantom{2}]{5}})+C$

B$)si{n}^{-1}(\frac{x}{5})+Csi{n}^{-1}(\frac{}{\sqrt[\phantom{2}]{5}})+C$

$q,$

If ${\displaystyle \underset{0}{\overset{1}{}}}f(x)+g(x)dx=10,{\displaystyle \underset{0}{\overset{1}{}}}f(x)=2,$ then ${\displaystyle \underset{0}{\overset{1}{}}}3g(x)+x^{2}dx=$

A$)\frac{25}{3}$

B$)24$

C $24\frac{1}{3}$

$7.$ The total area between the curve $y=1-x^2$ and the $x-$axis over the interval $0,2$

A$)\frac{8}{3}$

B$)2$

C$)\frac{7}{3}$

$8.$ The volume of the solid that results when the region enclosed by the curves $cosx,x=0,$ and $x=\frac{}{4}$ that revolved about the $x-$axis is

A$)\frac{}{2}$

B$)\frac{1}{2}$

C$)$

$q,$

$9.{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{dx}{\sqrt[\phantom{2}]{1+9x^2}}=$

A$)\frac{1}{3}cos{h}^{-1}(3x)+C$

r$)\frac{1}{3}sin{h}^{-1}(3x)+C$

C$)\frac{1}{3}tan{h}^{-1}(3x)$

$10.$ Using the integration by parts, ${\displaystyle \underset{0}{\overset{1}{}}}ta{n}^{-1}(x)dx=$

A$)\frac{}{4}-ln(2)$

D$)\frac{}{4}-\frac{ln(2)}{2}$

C$)\frac{}{4}-ln(4)$