第12类RD Sharma解–第19章不定积分–练习19.17

评估以下积分：

问题1.∫dx/√(2x – x ² )

解决方案：

We have,

Let I = ∫dx/√(2x – x²)

= ∫dx/√(1 – 1 + 2x – x²)

= ∫dx/√{1 – (x² – 2x + 1)}

= ∫dx/√{1² – (x – 1)²}

Let x – 1 = q …(1)

= ∫dx/√{1² – (q)²}

As we know that, ∫dx/√(a² – x²) = sin^-1(x/a)

So, 

= sin^-1(q) + C    

Now put the value of q from eq(1), we get

= sin^-1(x – 1) + C    

问题2。∫dx/√(8 + 3x – x ² )

解决方案：

We have,

Let I = ∫dx/√(8 + 3x – x²)

Here, (8 + 3x – x²) can be written as 8 – (x² – 3x + 9/4 – 9/4)

= (8 + 9/4) – (x – 3/2)²

= (41/4) – (x – 3/2)²

= $∫\frac{dx}{(\sqrt{\frac{41}{4}})^2-(x-\frac{3}{2})^2}$

Let x – 3/2 =  q …(1)

= $∫\frac{dx}{(\sqrt{\frac{41}{4}})^2-(q)^2}$

As we know that, ∫dx/√(a² – x²) = sin^-1(x/a)

So, 

= $sin^{-1}(\frac{q}{\sqrt{\frac{41}{4}}}) + C$   

Now put the value of q from eq(1), we get

= $sin^{-1}(\frac{x-\frac{3}{2}}{\frac{\sqrt{41}}{2}})+C$

= sin^-1(2x – 3/√41) + C

问题3.∫dx/√(5 – 4x – 2x ² )

解决方案：

We have,

Let I = ∫dx/√(5 – 4x – 2x²)

= ∫dx/√{2(5/2 – 2x – x²)}

= (1/√2)∫dx/√{5/2 – (x² – 2x + 1 – 1)}

= (1/√2)∫dx/√{(5/2 + 1) – (x² – 2x + 1)}

= (1/√2)∫dx/√{7/2 – (x – 1)²}

=$\frac{1}{\sqrt{2}}∫\frac{dx}{(\sqrt\frac{7}{2})^2-(x+1)^2}$

Let x + 1 = q …(1)

= $\frac{1}{\sqrt{2}}∫\frac{dx}{(\sqrt\frac{7}{2})^2-(q)^2}$

As we know that, ∫dx/√(a² – x²) = sin^-1(x/a)

So, 

= $\frac{1}{\sqrt{2}}sin^{-1}(\frac{q}{\sqrt\frac{7}{2}})+C$

Now put the value of q from eq(1), we get

= $\frac{1}{\sqrt{2}}sin^{-1}(\frac{x+1}{\sqrt\frac{7}{2}})+C$

= $\frac{1}{\sqrt{2}}sin^{-1}(\frac{(x+1)\sqrt{2}}{\sqrt7})+C$

问题4.∫dx/√(3x ² + 5x + 7)

解决方案：

We have,

Let I = ∫dx/√(3x² + 5x + 7)

= ∫dx/√{3(x² + 5x/3 + 7/3)}

= (1/√3)∫dx/√{5/2-(x2-2x+1-1)}

= $\frac{1}{\sqrt3}∫\frac{dx}{\sqrt{x^2+\frac{5x}{3}+(\frac{5}{6})^2-(\frac{5}{6})^2+\frac{7}{3}}}$

= $\frac{1}{\sqrt3}∫\frac{dx}{\sqrt{(x+\frac{5}{6})^2-\frac{25}{36}+\frac{7}{3}}}$

= $\frac{1}{\sqrt3}∫\frac{dx}{\sqrt{(x+\frac{5}{6})^2+(\frac{\sqrt{56}}{6})^2}}$

= $\frac{1}{\sqrt3}log|x+\frac{5}{6}+\sqrt{(x+\frac{5}{6})^2-(\frac{59}{36})}|+C$

= $\frac{1}{\sqrt3}log|x+\frac{5}{6}+\sqrt{x^2+\frac{5x}{3}+\frac{7}{3}}|+C$

问题5.∫dx/√{(x –α)(β– x)}

解决方案：

We have,

Let I = ∫dx/√{(x – α)(β – x)}

= ∫dx/√(-x² +αx + βx – αβ)

= ∫dx/√{-x² + x(α + β) – αβ}

= $∫\frac{dx}{\sqrt{-x^2+2x(\frac{α+β}{2})-(\frac{α+β}{2})^2+(\frac{α+β}{2})^2+αβ}}$

= $∫\frac{dx}{\sqrt{-[x-(\frac{α+β}{2})^2]+(\frac{α-β}{2})^2}}$

= $∫\frac{dx}{\sqrt{(\frac{α+β}{2})^2-[x-(\frac{α+β}{2})^2]}}$

Let x – (α + β)/2 = q …(1)

= $∫\frac{dx}{\sqrt{(\frac{α+β}{2})^2-(q)^2}}$

As we know that, ∫dx/√(a² – x²) = sin^-1(x/a)

So, 

= $sin^{-1}\frac{q}{\frac{α-β}{2}}+C$

Now put the value of q from eq(1), we get

= $sin^{-1}\frac{x-\frac{α+β}{2}}{\frac{α-β}{2}}+C$

= $sin^{-1}(\frac{2x-α-β}{α-β})+C$

问题6.∫dx/√(7 – 3x – 2x ² )

解决方案：

We have,

Let I = ∫dx/√(7 – 3x – 2x²)

= ∫dx/√{2(7/2 – 3x/2 – x²)}

= $\frac{1}{\sqrt{2}}∫\frac{dx}{\sqrt{\frac{7}{2}-[x^2+2x(\frac{3}{4})+(\frac{3}{4})^2-(\frac{3}{4})^2}]}$

= $\frac{1}{\sqrt{2}}∫\frac{dx}{\sqrt{\frac{7}{2}+\frac{9}{16}-(x+\frac{3}{4})^2}}$

= $\frac{1}{\sqrt{2}}∫\frac{dx}{\sqrt{\frac{65}{16}-(x+\frac{3}{4})^2}}$

= $\frac{1}{\sqrt{2}}∫\frac{dx}{\sqrt{(\frac{\sqrt{65}}{4})^2-(x+\frac{3}{4})^2}}$

Let x + 3/2 = q …(1)

= $\frac{1}{\sqrt{2}}∫\frac{dx}{\sqrt{(\frac{\sqrt{65}}{4})^2-(q)^2}}$

As we know that, ∫dx/√(a² – x²) = sin^-1(x/a)

So, 

= $\frac{1}{\sqrt{2}}sin^{-1}(\frac{q}{\frac{\sqrt{65}}{4}})+C$

Now put the value of q from eq(1), we get

= $\frac{1}{\sqrt{2}}sin^{-1}(\frac{x+\frac{3}{4}}{\frac{\sqrt{65}}{4}})+C$

= $\frac{1}{\sqrt{2}}sin^{-1}(\frac{4x+3}{\sqrt{65}})+C$

问题7.∫dx/√(16 – 6x – x ² )

解决方案：

We have,

Let I = ∫dx/√(16 – 6x – x²)

= ∫dx/√{16 – (x² + 2.3x + 9 – 9)}

= ∫dx/√{25 – (x² + 2.3x + 9)}

= $∫\frac{dx}{(5)^2-(x+3)^2}$

Let x + 3/2 = q …(1)

= $∫\frac{dx}{(5)^2-(q)^2}$

As we know that, ∫dx/√(a² – x²) = sin^-1(x/a)

So, 

= $sin^{-1}(\frac{q}{5})+C$

Now put the value of q from eq(1), we get

= $sin^{-1}(\frac{x+3}{5})+C$

问题8.∫dx/√(7 – 6x – x ² )

解决方案：

We have,

Let I = ∫dx/√(7 – 6x – x²)

= ∫dx/√{7-(x² + 2.3x + 9 – 9)}

= ∫dx/√{16 – (x² + 2.3x + 9)}

= $∫\frac{dx}{(4)^2-(x+3)^2}$

Let x + 3/2 = q …(1)

= $∫\frac{dx}{(4)^2-(q)^2}$

As we know that, ∫dx/√(a² – x²) = sin^-1(x/a)

So, 

= $sin^{-1}(\frac{q}{4})+C$

Now put the value of q from eq(1), we get

= $sin^{-1}(\frac{x+3}{4})+C$

问题9.∫dx/√(5x ² – 2x)

解决方案：

We have,

Let I = ∫dx/√(5x² – 2x)

= ∫dx/√{5(x² – 2x/5)}

= $\frac{1}{\sqrt5}∫\frac{dx}{\sqrt{x^2-\frac{2x}{5}+(\frac{1}{5})^2-(\frac{1}{5})^2}}$

= $\frac{1}{\sqrt5}∫\frac{dx}{\sqrt{(x-\frac{1}{5})^2-(\frac{1}{5})^2}}$

= $\frac{1}{\sqrt5}log|x-\frac{1}{5}+\sqrt{(x-\frac{1}{5})^2+(\frac{1}{5})^2}|+C$

= $\frac{1}{\sqrt5}log|x-\frac{1}{5}+\sqrt{x^2-\frac{2x}{5}}|+C$