KindiakMath

Several Laborious Integrals

Problem 1. Evaluate the integral \displaystyle \int \frac{(2x+1) \ln(9x+5)}{(3x+2)^4} \, \mathrm dx.

(Click for Solution)

Solution. We first evaluate \displaystyle \int \frac{(2x+1) }{(3x+2)^4} \, \mathrm dx for simplicity. Integrating by parts,

\displaystyle \begin{aligned} \int \frac{(2x+1) }{(3x+2)^4} \, \mathrm dx &= \underbrace{ \frac{(3x+2)^{-3}}{-3 \cdot 3} }_{\mathrm I} \cdot \underbrace{(2x+1)}_{\mathrm S} - \int \underbrace{ \frac{(3x+2)^{-3}}{-3 \cdot 3} }_{\mathrm I} \cdot \underbrace{2}_{\mathrm D}\, \mathrm dx \\ &= -\frac{2x+1}{9(3x+2)^3} + \frac 29 \int (3x+2)^{-3}\, \mathrm dx \\ &= -\frac{2x+1}{9(3x+2)^3} + \frac 29 \cdot \frac{(3x+2)^{-2}}{-2 \cdot 3} +C \\ &= -\frac{2x+1}{9(3x+2)^3} - \frac 1{27(3x+2)^2} +C \\ &= -\frac{3(2x+1)}{27(3x+2)^3} - \frac {3x+2}{27(3x+2)^3} +C \\ &= -\frac{3(2x+1) + 3x+2}{27(3x+2)^3} +C \\ &= -\frac{9x+5}{27(3x+2)^3} + C.\end{aligned}

Therefore, integrating by parts again,

\displaystyle \begin{aligned} & \int \frac{(2x+1) \ln(9x+5)}{(3x+2)^4} \, \mathrm dx \\ &= \underbrace{-\frac{9x+5}{27(3x+2)^3}}_{\mathrm I} \cdot \underbrace{\ln(9x+5)}_{\mathrm S} - \int \underbrace{-\frac{9x+5}{27(3x+2)^3}}_{\mathrm I}\cdot \underbrace{\frac 1{9x+5} \cdot 9}_{\mathrm D}\, \mathrm dx \\ &= -\frac{(9x+5)\ln(9x+5)}{27(3x+2)^3} + \frac 13 \int \frac{1}{(3x+2)^3}\, \mathrm dx \\ &= -\frac{(9x+5)\ln(9x+5)}{27(3x+2)^3} + \frac 13 \int (3x+2)^{-3}\, \mathrm dx \\ &= -\frac{(9x+5)\ln(9x+5)}{27(3x+2)^3} + \frac 13 \cdot \frac 13 \cdot \frac{(3x+2)^{-2}}{-2} + C \\ &= -\frac{(9x+5)\ln(9x+5)}{27(3x+2)^3} - \frac 1{18 (3x+2)^2} + C.\end{aligned}

Problem 2. Evaluate the integral \displaystyle \int_4^5 \frac{9x^3 - 14x^2 + 8x - 11}{(x-1)(x-2)(x^2+3)} \, \mathrm dx.

(Click for Solution)

Solution. We decompose into a sum of partial fractions as follows:

\displaystyle \frac{9x^3 - 14x^2 + 8x - 11}{(x-1)(x-2)(x^2+3)} = \frac{A}{x-1} + \frac{B}{x-2} + \frac{Cx+D}{x^2+3}.

Assuming we have evaluated A,B,C,D, integrating yields

\displaystyle \begin{aligned} &\int_4^5 \frac{9x^3 - 14x^2 + 8x - 11}{(x-1)(x-2)(x^2+3)} \, \mathrm dx \\ &= \int_4^5 \left(A \cdot \frac{1}{x-1} + B \cdot \frac{1}{x-2} + \frac{C}{2} \cdot \frac{2x}{x^2+3} + D \cdot \frac{1}{(\sqrt 3)^2 + x^2}\right)\, \mathrm dx \\ &= \left[A \ln|x-1| + B \ln |x-2| + \frac C2 \ln(x^2+3) + \frac{D}{\sqrt 3} \tan^{-1}\left( \frac{x}{\sqrt 3}\right)\right]_4^5 \\ &= A \ln \frac 43 + B \ln \frac 32 + \frac C2 \ln \frac{28}{19} + \frac{D}{\sqrt 3} \left( \tan^{-1}\frac{5}{\sqrt 3} - \tan^{-1}\frac{4}{\sqrt 3}\right). \end{aligned}

It remains to evaluate A,B,C,D. Cross-multiplying yields

\displaystyle \begin{aligned} 9x^3 - 14x^2 + 8x - 11 &= A(x-2)(x^2+3) \\ &\phantom{=} + B(x-1)(x^2+3)\\ &\phantom{=} + (Cx+D)(x-1)(x-2).\end{aligned}

Setting x = 1 yields A = 2.
Setting x = 2 yields B = 3.

Comparing the terms involving x^3,

9 = A + B + C \quad \Rightarrow \quad C = 4.

Comparing the constant terms,

-11 = -6A + (-3B) + 2D \quad \Rightarrow \quad D = 5.

Therefore, the definite integral evaluates to

\begin{aligned} &\int_4^5 \frac{9x^3 - 14x^2 + 8x - 11}{(x-1)(x-2)(x^2+3)} \, \mathrm dx \\ &=2 \ln \frac 43 + 3 \ln \frac 32 + \frac 42 \ln \frac{28}{19} + \frac{5}{\sqrt 3} \left( \tan^{-1}\frac{5}{\sqrt 3} - \tan^{-1}\frac{4}{\sqrt 3}\right)\\ &= 2 \ln \frac {112}{57} + 3 \ln \frac 32 + \frac{5}{\sqrt 3} \left( \tan^{-1}\frac{5}{\sqrt 3} - \tan^{-1}\frac{4}{\sqrt 3}\right) \\ &= 2 \ln \frac {112}{57} + 3 \ln \frac 32 + \frac{5}{\sqrt 3}\tan^{-1}\frac{\sqrt 3}{23},\end{aligned}

where the last equality follows from the identity

\displaystyle \tan^{-1}(a) - \tan^{-1}(b) = \tan^{-1} \left(\frac{a-b}{1+ab}\right).

Problem 3. Given a > 0 and b^2 - 4ac < 0, evaluate \displaystyle \int \frac{1}{\sqrt{ax^2 + bx + c}}\, \mathrm dx. Deduce the value of

\displaystyle \lim_{n \to \infty} \left( \frac{1}{\sqrt{n^2+1^2}} + \frac{1}{\sqrt{n^2+2^2}} + \cdots + \frac{1}{\sqrt{n^2+n^2}} \right).

(Click for Solution)

Solution. For the integral, we first complete the square

ax^2 + bx + c = a(x+h)^2 + k,

where h and k are constants to be determined in terms of a,b,c. To determine h, k, we expand the right-hand side to obtain

ax^2 + bx + c = ax^2 + (2ah)x + ah^2 + k.

Comparing coefficients h = b/(2a) and

\begin{aligned} k = c - ah^2 &= c - a \left(\frac b{2a}\right)^2 \\ &= c - a \cdot \frac{b^2}{4a^2} \\ &= \frac{4ac- b^2}{4a} \\ &= \frac{-(b^2 - 4ac)}{4a} > 0. \end{aligned}

Therefore,

\begin{aligned} &\int \frac{1}{\sqrt{ax^2 + bx + c}}\, \mathrm dx \\ &= \int \frac{1}{\sqrt{a(x+h)^2 + k}}\, \mathrm dx \\ &= \frac{1}{\sqrt a} \int \frac{1}{\sqrt{(x+h)^2 + (\sqrt{k/a})^2 }}\, \mathrm dx \\ &= \frac 1{\sqrt a} \ln \left|(x+h) + \sqrt{(x+h)^2 + (\sqrt{k/a})^2} \right| + C \\ &= \frac 1{\sqrt a} \ln \left| x+\frac b{2a} + \sqrt{ \left(x+\frac b{2a} \right)^2 + \frac{4ac-b^2}{4a^2} } \right| + C. \end{aligned}

For the limit, we factorise n^2 to obtain

\displaystyle \begin{aligned} &\lim_{n \to \infty} \left( \frac{1}{\sqrt{n^2+1^2}} + \frac{1}{\sqrt{n^2+2^2}} + \cdots + \frac{1}{\sqrt{n^2+n^2}} \right) \\ &=\lim_{n \to \infty} \frac 1n \left( \frac{1}{\sqrt{1+\left( \frac 1n\right)^2}} + \frac{1}{\sqrt{1+\left( \frac 2n\right)^2}} + \cdots + \frac{1}{\sqrt{1+\left( \frac nn\right)^2}} \right) \\ &= \int_0^1 \frac{1}{\sqrt{1+x^2}}\, \mathrm dx, \end{aligned}

where the last equality follows from the definition of integration. Setting a = 1, b = 0, c = 1 in the indefinite integral,

\begin{aligned} \int_0^1 \frac{1}{\sqrt{1+x^2}}\, \mathrm dx &= \left[ \frac 1{\sqrt 1} \ln \left| x+ 0 + \sqrt{ (x+ 0 )^2 + \frac{4 \cdot 1 \cdot 1 - 0^2}{4 \cdot 1^2} } \right| \right]_0^1 \\ &= \left[ \ln \left| x + \sqrt{ x^2 + 1 } \right| \right]_0^1 \\ &= \ln ( 1 + \sqrt{ 2 } ) - 0 \\ &= \ln ( 1 + \sqrt{ 2 } ).\end{aligned}

—Joel Kindiak, 24 Feb 25, 1800H

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