Oblicz całki

[Sway22]

a) \( \int_{}^{} \frac{x+3}{x^2+4x+8} dx \)
b) \( \int_{0}^{ \pi } x^2 \cos x dx\)
c) \( \int_{}^{} \frac{dx}{x^3+x^2+x} dx\)

[eresh]

a) \( \int_{}^{} \frac{x+3}{x^2+4x+8} dx \)

\(\int\frac{x+3}{x^2+4x+8}dx=\frac{1}{2}\int\frac{2x+4+1}{x^2+4x+8}dx=\frac{1}{2}\int\frac{2x+4}{x^2+4x+8}dx+\frac{1}{2}\int\frac{dx}{(x+2)^2+4}=\frac{1}{2}\ln(x^2+4x+8)+\frac{1}{2}\arctg\frac{x+2}{2}+C\)

[eresh]

b) \( \int_{0}^{ \pi } x^2 \cos x dx\)

\(\int x^2\cos xdx= \begin{bmatrix}u(x)=x^2&u'(x)=2x\\v'(x)=\cos x&v(x)=\sin x \end{bmatrix}=x^2\sin x-2\int x\sin xdx= \\=\begin{bmatrix}u(x)=x&u'(x)=1\\v'(x)=\sin x&v(x)=-\cos x \end{bmatrix}=x^2\sin x+2x\cos x-2\int\cos xdx=x^2\sin x+2x\cos x-2\sin x+C\)

\(\int_{0}^{ \pi } x^2 \cos x dx=-2\pi\)

[Jerry]

a) \( \int \frac{x+3}{x^2+4x+8} dx \)

\(\int \frac{x+3}{x^2+4x+8} dx={1\over2}\cdot\int \frac{2x+4}{x^2+4x+8} dx+\int \frac{1}{(x+2)^2+4} dx\)

• \(I_1=\ln(x^2+4x+8)+C_1\)
• \(I_2={1\over2}\cdot\arctg\frac{x+2}{2}+C_2\)
  bo dla \(x+2=2t\) mamy \(dx=2dt\)

Pozdrawiam

[edited] zwłaszcza eresh

[eresh]

c) \( \int_{}^{} \frac{dx}{x^3+x^2+x} dx\)

\(\frac{1}{x(x^2+x+1)}=\frac{A}{x}+\frac{Bx+C}{x^2+x+1}\\
1=Ax^2+Ax+A+Bx^2+Cx\\
\begin{cases}A+B=0\\A+C=0\\A=1\end{cases}\\
\begin{cases}A=1\\B=-1\\C=-1\end{cases}\\
\int\frac{dx}{x^3+x^2+x}=\int\frac{dx}{x}-\int\frac{x+1}{x^2+x+1}=\ln|x|-\frac{1}{2}\int\frac{2x+1+1}{x^2+x+1}=\ln |x|-\frac{1}{2}\ln|x^2+x+1|-\frac{1}{2}\int\frac{dx}{(x+0,5)^2+\frac{3}{4}}=\\=\ln |x|-\frac{1}{2}\ln|x^2+x+1|-\frac{1}{2}\arctg\frac{x+0,5}{\frac{\sqrt{3}}{2}}+C\)