Question

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question is in picture

R=4

Answer 1

a)

$$\int{x^2 \over (x+3)(2x-3)^2}dx$$ $${x^2 \over (x+3)(2x-3)^2}={A \over x+3}+{B\over (2x-3)^2}+{D\over 2x-3}=$$

$$={4Ax^2-12Ax+9A+Bx+3B+2Dx^2+3Dx-9D\over (x+3)(2x-3)^2}$$

$4A+2D=1,-12+B+3D=0,9A+3B-9D=0$

$A=\dfrac{1}{9},B=\dfrac{1}{2}, D=\dfrac{5}{18}$

$$\int{x^2 \over (x+3)(2x-3)^2}dx={1 \over 9}\int{dx \over x+3}+{1 \over 2}\int{dx \over (2x-3)^2}+{5 \over 18}\int{dx \over 2x-3}=$$

$$={1 \over 9}\ln|x+3|+{5\over 36}\ln|2x-3|-{1 \over 4(2x-3)}+C$$

b)

$$\int{6-5x \over (2x+1)(x^2+4)}dx$$

$${6-5x \over (2x+1)(x^2+4)}={Ax+B \over x^2+4}+{D\over 2x+1}=$$

$$={2Ax^2+Ax+2Bx+B+Dx^2+4D\over (2x+1)(x^2+4)}$$

$2A+D=0, A+2B=-5,B+4D=6$

$A=-1,B=-2,D=2$

$$\int{6-5x \over (2x+1)(x^2+4)}dx=-\int{x\over x^2+4}dx-\int{2\over x^2+4}dx+\int{2\over 2x+1}dx=$$

$$=\ln|2x+1|-{1 \over 2}\ln(x^2+4)-\arctan({x \over 2})+C$$

c)

$$\int x^2\sin(2x-1)dx$$

$$\int udv=uv-\int vdu$$

$$u=x^2, du=2xdx$$ $$dv=\sin(2x-1)dx, v=\int \sin(2x-1)dx=-{1 \over 2}\cos(2x-1)$$

$$\int x^2\sin(2x-1)dx=-{1 \over 2}x^2\cos(2x-1)+\int x\cos(2x-1)$$

$$\int udv=uv-\int vdu$$

$$u=x, du=dx$$

$$dv=\cos(2x-1)dx, v=\int \cos(2x-1)dx={1 \over 2}\sin(2x-1)$$

$$\int x^2\sin(2x-1)dx=$$ $$=-{1 \over 2}x^2\cos(2x-1)+{1 \over 2}\sin(2x-1)-{1 \over 2}\int \sin(2x-1)dx=$$

$$=-{1 \over 2}x^2\cos(2x-1)+{1 \over 2}\sin(2x-1)+{1 \over 4}\cos(2x-1)+C$$

d)

$$\int \sin^{-1}(6x)dx$$ $$\int udv=uv-\int vdu$$ $$u= \sin^{-1}(6x), du={6 \over \sqrt{1-36x^2}}dx$$ $$dv=dx, v=x$$

$$\int \sin^{-1}(6x)dx=x\sin^{-1}(6x)-6\int {x \over \sqrt{1-36x^2}}dx=$$

$$=x\sin^{-1}(6x)+{1 \over 6}\sqrt{1-36x^2}+C$$

e)

$$\int {1 \over (e^x-e^{-x})^2}dx$$

$$u=e^{2x}-1, du=2e^{2x}dx$$

$$\int {1 \over (e^x-e^{-x})^2}dx=\int {e^{2x} \over (e^{2x}-1)^2}dx={1\over 2}\int {du \over u^2}=$$

$$=-{1 \over 2u}+C=-{1 \over 2(e^{2x}-1)}+C$$

f)

$$\int{1 \over x(x^2+7)}dx=\int{1 \over x^3(1+{7 \over x^2})}dx$$

$u=1+\dfrac{7}{x^2},du=-\dfrac{14}{x^3}dx$

$$\int{1 \over x(x^2+7)}dx=\int{1 \over x^3(1+{7 \over x^2})}dx=$$

$$=-{1 \over 14}\int{du \over u}=-{1 \over 14}\ln|u|+C=-{1 \over 14}\ln(1+{7 \over x^2})+C$$

$$\int{1 \over x(x^2+7)}dx=-{1 \over 14}\ln(1+{7 \over x^2})+C$$

g)

$$\int (1-x)^6(1+x)^2dx$$ $$u=x-1, du=dx$$

$$\int (1-x)^6(1+x)^2dx=\int u^6(u+2)^2du=$$ $$=\int(u^8+4u^7+4u^6)du={1 \over9}u^9+{1 \over 2}u^8+{4 \over 7}u^7+C=$$ $$={1 \over9}(x-1)^9+{1 \over 2}(x-1)^8+{4 \over 7}(x-1)^7+C$$

h)

$$\int {1 \over x^3(x^2+1)}dx=\int{1 \over x^5(1+{1 \over x^2})}dx$$

$u=1+\dfrac{1}{x^2},du=-\dfrac{2}{x^3}dx$

$$\int {1 \over x^5(x^2+1)}dx=\int{1 \over x^3(1+{1 \over x^2})}dx=$$

$$=-{1 \over 2}\int{u-1 \over u}du=-{1 \over 2}u+{1 \over 2}\ln|u|+C=$$

$$=-{1 \over 2}(1+{1 \over x^2})+{1 \over 2}\ln(1+{1 \over x^2})+C=$$

$$=-{1 \over 2x^2}+{1 \over 2}\ln(1+{1 \over x^2})+C$$