Question

Ma poate ajuta cineva cu rezolvările?​

Answer 1

Răspuns:

Explicație pas cu pas:

1.a.

$u=x^2+1= > du=(2x+1)dx\\\int(x^2+1)^3*2xdx=\int u^3du=\frac{u^4}{4}= \frac{(x^2+1)^4}{4}$

$= > \int(x^2+1)^3*2xdx= \frac{(x^2+1)^4}{4}+C$

b.

$u=sin(x)= > du=cos(x)dx\\\int sin^3(x)cos(x)dx=\int u^3du=\frac{u^4}{4}= \frac{sin^4(x)}{4}$

$= > \int sin^3(x)cos(x)dx= \frac{sin^4(x)}{4}$

2.a.

$u=x^2+x+2009= > du=(2x+1)dx\\\int \frac{2x+1}{x^2+x+2009}dx =\int \frac{1}{u}du=ln(u) =ln(x^2+x+2009)$

$= > \int \frac{2x+1}{x^2+x+2009}dx =ln(x^2+x+2009)+C$

b.

$u=cos(x)= > du=-sin(x)dx= > sin(x)dx=-du\\\int \frac{sin(x)}{7cos(x)}=-\frac{1}{7} \int \frac{1}{u} =-\frac{1}{7}ln(u)=-\frac{1}{7}ln(cos(x))$

$= > \int \frac{sin(x)}{7cos(x)}=-\frac{ln(|cos(x)|)}{7}+C$

3.a.

$\int\limits^1_0 {3x^2+4x^3+11} \, dx=(\frac{4x^4}{4} +\frac{3x^3}{3} +11x)^1_0=(x^4+x^3+11)^1_0=(1+1+11)-(0+0+0)=13$

b.

$\int\limits^\pi_0 {sin(x)-3cos(x)} \, dx= (-cos(x)-3sin(x))^\pi_0=(-cos(\pi )-3sin(\pi ))-(-cos(0)-3sin(0))=(-(-1)-0)-(-1-0)=1+1=2$

c.

$\int\limits^4_1 {\sqrt{x}+x^\frac{3}{2}} \, dx=(\frac{2x^\frac{3}{2} }{3}+\frac{2x\frac{5}{2} }{5} )^4_1=(\frac{2x^\frac{3}{2}(3x+5)}{15})^4_1 =(\frac{2*4^\frac{3}{2}(3*4+5)}{15})-(\frac{2*1^\frac{3}{2}(3*1+5)}{15})=\frac{272-16}{15} =\frac{256}{15}$

d.

$u=x^2+3= > du=2xdx= > \frac{du}{2}=xdx \\\int\ {\frac{x}{x^2+3}}\, dx= \frac{1}{2} \int \frac{1}{u}du=\frac{ln(u)}{2}=\frac{ln(x^2+3)}{2}+C \\\int\limits^2_1 {\frac{x}{x^2+3}}\, dx=(\frac{ln(x^2+3)}{2} )^2_1=(\frac{ln(2^2+3)}{2} )-(\frac{ln(1^2+3)}{2} )=\frac{ln(7)-ln(4)}{2}=\frac{ln(\frac{7}{4} )}{2}$