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Question: $\frac{1}{x^3(1+x)^{3/2}}.$...

1x3(1+x)3/2.\frac{1}{x^3(1+x)^{3/2}}.

Answer

15x2+5x−24x21+x+158ln⁡∣1+x−11+x+1∣+C\frac{15x^2+5x-2}{4x^2\sqrt{1+x}} + \frac{15}{8}\ln\left|\frac{\sqrt{1+x}-1}{\sqrt{1+x}+1}\right| + C

Explanation

Solution

To evaluate the integral I=∫1x3(1+x)3/2dxI = \int \frac{1}{x^3(1+x)^{3/2}} dx, we use the substitution method.

The integrand is of the form ∫xm(a+bxn)pdx\int x^m (a+bx^n)^p dx. Here, m=−3m=-3, a=1a=1, b=1b=1, n=1n=1, and p=−3/2p=-3/2. Since m+1n=−3+11=−2\frac{m+1}{n} = \frac{-3+1}{1} = -2, which is an integer, the appropriate substitution is a+bxn=tka+bx^n = t^k, where kk is the denominator of pp. So, we let 1+x=t21+x = t^2.

From 1+x=t21+x = t^2:

  1. x=t2−1x = t^2-1
  2. dx=2t dtdx = 2t \, dt
  3. (1+x)3/2=(t2)3/2=t3(1+x)^{3/2} = (t^2)^{3/2} = t^3
  4. x3=(t2−1)3x^3 = (t^2-1)^3

Substitute these into the integral: I=∫1(t2−1)3⋅t3⋅(2t dt)I = \int \frac{1}{(t^2-1)^3 \cdot t^3} \cdot (2t \, dt) I=∫2tt3(t2−1)3dtI = \int \frac{2t}{t^3(t^2-1)^3} dt I=∫2t2(t2−1)3dtI = \int \frac{2}{t^2(t^2-1)^3} dt

Now, we use partial fraction decomposition for the integrand 2t2(t2−1)3\frac{2}{t^2(t^2-1)^3}. Let y=t2y = t^2. Then the expression becomes 2y(y−1)3\frac{2}{y(y-1)^3}. We decompose this into partial fractions: 2y(y−1)3=Ay+By−1+C(y−1)2+D(y−1)3\frac{2}{y(y-1)^3} = \frac{A}{y} + \frac{B}{y-1} + \frac{C}{(y-1)^2} + \frac{D}{(y-1)^3} To find the coefficients A, B, C, D: Multiply both sides by y(y−1)3y(y-1)^3: 2=A(y−1)3+By(y−1)2+Cy(y−1)+Dy2 = A(y-1)^3 + By(y-1)^2 + Cy(y-1) + Dy

Set y=1y=1: 2=D(1)  ⟹  D=22 = D(1) \implies D=2. Set y=0y=0: 2=A(−1)3  ⟹  A=−22 = A(-1)^3 \implies A=-2.

Now, substitute A=−2A=-2 and D=2D=2 back into the equation: 2=−2(y−1)3+By(y−1)2+Cy(y−1)+2y2 = -2(y-1)^3 + By(y-1)^2 + Cy(y-1) + 2y 2=−2(y3−3y2+3y−1)+B(y3−2y2+y)+C(y2−y)+2y2 = -2(y^3-3y^2+3y-1) + B(y^3-2y^2+y) + C(y^2-y) + 2y 2=−2y3+6y2−6y+2+By3−2By2+By+Cy2−Cy+2y2 = -2y^3+6y^2-6y+2 + By^3-2By^2+By + Cy^2-Cy + 2y

Equating coefficients of powers of yy: Coefficient of y3y^3: 0=−2+B  ⟹  B=20 = -2+B \implies B=2. Coefficient of y2y^2: 0=6−2B+C  ⟹  0=6−2(2)+C  ⟹  0=2+C  ⟹  C=−20 = 6-2B+C \implies 0 = 6-2(2)+C \implies 0=2+C \implies C=-2.

So, the partial fraction decomposition is: 2y(y−1)3=−2y+2y−1−2(y−1)2+2(y−1)3\frac{2}{y(y-1)^3} = -\frac{2}{y} + \frac{2}{y-1} - \frac{2}{(y-1)^2} + \frac{2}{(y-1)^3}

Substitute back y=t2y=t^2: I=∫(−2t2+2t2−1−2(t2−1)2+2(t2−1)3)dtI = \int \left(-\frac{2}{t^2} + \frac{2}{t^2-1} - \frac{2}{(t^2-1)^2} + \frac{2}{(t^2-1)^3}\right) dt

Now, integrate each term:

  1. ∫−2t2dt=−2∫t−2dt=−2(t−1−1)=2t\int -\frac{2}{t^2} dt = -2 \int t^{-2} dt = -2 \left(\frac{t^{-1}}{-1}\right) = \frac{2}{t}

  2. ∫2t2−1dt=2∫1(t−1)(t+1)dt=2∫12(1t−1−1t+1)dt\int \frac{2}{t^2-1} dt = 2 \int \frac{1}{(t-1)(t+1)} dt = 2 \int \frac{1}{2}\left(\frac{1}{t-1} - \frac{1}{t+1}\right) dt =∫(1t−1−1t+1)dt=ln⁡∣t−1∣−ln⁡∣t+1∣=ln⁡∣t−1t+1∣= \int \left(\frac{1}{t-1} - \frac{1}{t+1}\right) dt = \ln|t-1| - \ln|t+1| = \ln\left|\frac{t-1}{t+1}\right|

  3. ∫−2(t2−1)2dt=−2∫1(t2−1)2dt\int -\frac{2}{(t^2-1)^2} dt = -2 \int \frac{1}{(t^2-1)^2} dt Use the reduction formula ∫1(x2−a2)ndx=−x2a2(n−1)(x2−a2)n−1−2n−32a2(n−1)∫1(x2−a2)n−1dx\int \frac{1}{(x^2-a^2)^n} dx = \frac{-x}{2a^2(n-1)(x^2-a^2)^{n-1}} - \frac{2n-3}{2a^2(n-1)} \int \frac{1}{(x^2-a^2)^{n-1}} dx. For n=2,a=1n=2, a=1: ∫1(t2−1)2dt=−t2(1)(1)(t2−1)1−2(2)−32(1)(1)∫1(t2−1)1dt\int \frac{1}{(t^2-1)^2} dt = \frac{-t}{2(1)(1)(t^2-1)^1} - \frac{2(2)-3}{2(1)(1)} \int \frac{1}{(t^2-1)^1} dt =−t2(t2−1)−12∫1(t−1)(t+1)dt= \frac{-t}{2(t^2-1)} - \frac{1}{2} \int \frac{1}{(t-1)(t+1)} dt =−t2(t2−1)−12⋅12ln⁡∣t−1t+1∣=−t2(t2−1)−14ln⁡∣t−1t+1∣= \frac{-t}{2(t^2-1)} - \frac{1}{2} \cdot \frac{1}{2} \ln\left|\frac{t-1}{t+1}\right| = \frac{-t}{2(t^2-1)} - \frac{1}{4}\ln\left|\frac{t-1}{t+1}\right|. So, −2∫1(t2−1)2dt=−2(−t2(t2−1)−14ln⁡∣t−1t+1∣)=tt2−1+12ln⁡∣t−1t+1∣-2 \int \frac{1}{(t^2-1)^2} dt = -2 \left(\frac{-t}{2(t^2-1)} - \frac{1}{4}\ln\left|\frac{t-1}{t+1}\right|\right) = \frac{t}{t^2-1} + \frac{1}{2}\ln\left|\frac{t-1}{t+1}\right|.

  4. ∫2(t2−1)3dt=2∫1(t2−1)3dt\int \frac{2}{(t^2-1)^3} dt = 2 \int \frac{1}{(t^2-1)^3} dt Using the reduction formula for n=3,a=1n=3, a=1: ∫1(t2−1)3dt=−t2(1)(2)(t2−1)2−2(3)−32(1)(2)∫1(t2−1)2dt\int \frac{1}{(t^2-1)^3} dt = \frac{-t}{2(1)(2)(t^2-1)^2} - \frac{2(3)-3}{2(1)(2)} \int \frac{1}{(t^2-1)^2} dt =−t4(t2−1)2−34∫1(t2−1)2dt= \frac{-t}{4(t^2-1)^2} - \frac{3}{4} \int \frac{1}{(t^2-1)^2} dt Substitute the result from step 3 for ∫1(t2−1)2dt\int \frac{1}{(t^2-1)^2} dt: =−t4(t2−1)2−34(−t2(t2−1)−14ln⁡∣t−1t+1∣)= \frac{-t}{4(t^2-1)^2} - \frac{3}{4} \left(\frac{-t}{2(t^2-1)} - \frac{1}{4}\ln\left|\frac{t-1}{t+1}\right|\right) =−t4(t2−1)2+3t8(t2−1)+316ln⁡∣t−1t+1∣= \frac{-t}{4(t^2-1)^2} + \frac{3t}{8(t^2-1)} + \frac{3}{16}\ln\left|\frac{t-1}{t+1}\right|. So, 2∫1(t2−1)3dt=2(−t4(t2−1)2+3t8(t2−1)+316ln⁡∣t−1t+1∣)2 \int \frac{1}{(t^2-1)^3} dt = 2 \left(\frac{-t}{4(t^2-1)^2} + \frac{3t}{8(t^2-1)} + \frac{3}{16}\ln\left|\frac{t-1}{t+1}\right|\right) =−t2(t2−1)2+3t4(t2−1)+38ln⁡∣t−1t+1∣= \frac{-t}{2(t^2-1)^2} + \frac{3t}{4(t^2-1)} + \frac{3}{8}\ln\left|\frac{t-1}{t+1}\right|.

Now, sum all the integrated terms: I=2t+ln⁡∣t−1t+1∣+(tt2−1+12ln⁡∣t−1t+1∣)+(−t2(t2−1)2+3t4(t2−1)+38ln⁡∣t−1t+1∣)+CI = \frac{2}{t} + \ln\left|\frac{t-1}{t+1}\right| + \left(\frac{t}{t^2-1} + \frac{1}{2}\ln\left|\frac{t-1}{t+1}\right|\right) + \left(\frac{-t}{2(t^2-1)^2} + \frac{3t}{4(t^2-1)} + \frac{3}{8}\ln\left|\frac{t-1}{t+1}\right|\right) + C

Combine logarithmic terms: (1+12+38)ln⁡∣t−1t+1∣=(8+4+38)ln⁡∣t−1t+1∣=158ln⁡∣t−1t+1∣\left(1 + \frac{1}{2} + \frac{3}{8}\right)\ln\left|\frac{t-1}{t+1}\right| = \left(\frac{8+4+3}{8}\right)\ln\left|\frac{t-1}{t+1}\right| = \frac{15}{8}\ln\left|\frac{t-1}{t+1}\right|.

Combine terms with tt2−1\frac{t}{t^2-1}: (1+34)tt2−1=74tt2−1\left(1 + \frac{3}{4}\right)\frac{t}{t^2-1} = \frac{7}{4}\frac{t}{t^2-1}.

So, the integral is: I=2t+7t4(t2−1)−t2(t2−1)2+158ln⁡∣t−1t+1∣+CI = \frac{2}{t} + \frac{7t}{4(t^2-1)} - \frac{t}{2(t^2-1)^2} + \frac{15}{8}\ln\left|\frac{t-1}{t+1}\right| + C

Substitute back t=1+xt = \sqrt{1+x} and t2−1=xt^2-1 = x: I=21+x+71+x4x−1+x2x2+158ln⁡∣1+x−11+x+1∣+CI = \frac{2}{\sqrt{1+x}} + \frac{7\sqrt{1+x}}{4x} - \frac{\sqrt{1+x}}{2x^2} + \frac{15}{8}\ln\left|\frac{\sqrt{1+x}-1}{\sqrt{1+x}+1}\right| + C

Combine the algebraic terms: 21+x+71+x4x−1+x2x2\frac{2}{\sqrt{1+x}} + \frac{7\sqrt{1+x}}{4x} - \frac{\sqrt{1+x}}{2x^2} Find a common denominator, which is 4x21+x4x^2\sqrt{1+x}: =2⋅4x24x21+x+71+x⋅x1+x4x21+x−1+x⋅21+x4x21+x= \frac{2 \cdot 4x^2}{4x^2\sqrt{1+x}} + \frac{7\sqrt{1+x} \cdot x\sqrt{1+x}}{4x^2\sqrt{1+x}} - \frac{\sqrt{1+x} \cdot 2\sqrt{1+x}}{4x^2\sqrt{1+x}} =8x2+7x(1+x)−2(1+x)4x21+x= \frac{8x^2 + 7x(1+x) - 2(1+x)}{4x^2\sqrt{1+x}} =8x2+7x+7x2−2−2x4x21+x= \frac{8x^2 + 7x + 7x^2 - 2 - 2x}{4x^2\sqrt{1+x}} =15x2+5x−24x21+x= \frac{15x^2 + 5x - 2}{4x^2\sqrt{1+x}}

Thus, the final integral is: I=15x2+5x−24x21+x+158ln⁡∣1+x−11+x+1∣+CI = \frac{15x^2 + 5x - 2}{4x^2\sqrt{1+x}} + \frac{15}{8}\ln\left|\frac{\sqrt{1+x}-1}{\sqrt{1+x}+1}\right| + C