/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 5 Evaluate the following integrals... [FREE SOLUTION] | 91影视

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Chapter 1: Problem 5

Evaluate the following integrals: (i) \(\int \frac{x^{7}}{\left(x^{12}-1\right)} d x\) (ii) \(\int \frac{x^{9} d x}{\left(x^{4}-1\right)^{2}}\)

Short Answer

Expert verified
Question: Evaluate the following integrals: (i) \(\int \frac{x^{7}}{\left(x^{12}-1\right)} d x\) and (ii) \(\int \frac{x^{9} d x}{\left(x^{4}-1\right)^{2}}\) Answer: (i) \(\frac{1}{12} \left(\ln \left|\frac{x^6 - 1}{x^6 + 1}\right| + C \right)\) (ii) \(\frac{1}{16} \left(\ln |x^4 - 1| - \frac{1}{x^4 - 1} + K \right)\)

Step by step solution

01

Substitution

Let \(u = x^6\), so \(u^2 = x^{12}\) and \(du = 6x^5 dx\). We can substitute to obtain: \(\frac{1}{6} \int \frac{u du}{\left(u^2 - 1\right)}\)
02

Partial Fractions

Now we will decompose the expression \(\frac{u}{u^2 - 1}\) into partial fractions: \(\frac{u}{u^2 - 1} = \frac{A}{u - 1} + \frac{B}{u + 1}\) Multiplying both sides by \(u^2 - 1\), we have: \(u = A(u + 1) + B(u - 1)\) By comparing the coefficients or plugging in suitable values for \(u\), we find \(A = \frac{1}{2}\) and \(B = -\frac{1}{2}\).
03

Integration

Now we can rewrite and integrate the expression: \(\frac{1}{6} \int \frac{u du}{\left(u^2 - 1\right)} = \frac{1}{6} \int \left(\frac{1}{2} \frac{du}{u - 1} - \frac{1}{2} \frac{du}{u + 1} \right)\) Integrating each term: \(\frac{1}{6} \left(\frac{1}{2} \ln |u - 1| - \frac{1}{2} \ln |u + 1| + C \right)\)
04

Reverse Substitution

Now we substitute back to get the result in terms of \(x\): \(\frac{1}{12} \left(\ln |x^6 - 1| - \ln |x^6 + 1| + C \right)\) The final result of the integral (i) is: \(\frac{1}{12} \left(\ln \left|\frac{x^6 - 1}{x^6 + 1}\right| + C \right)\) (ii) \(\int \frac{x^{9} d x}{\left(x^{4}-1\right)^{2}}\)
05

Substitution

Let \(v = x^4\), so \(v^2 = x^8\) and \(dv = 4x^3 dx\). We can substitute to obtain: \(\frac{1}{4} \int \frac{v^2 x dx}{\left(v - 1\right)^{2}}\)
06

Partial Fractions

Now we will decompose the expression \(\frac{v^2}{\left(v - 1\right)^2}\) into partial fractions: \(\frac{v^2}{(v - 1)^2} = \frac{C}{v - 1} + \frac{D}{(v - 1)^2}\) Multiplying both sides by \((v - 1)^2\), we have: \(v^2 = C(v - 1) + D\) By comparing the coefficients or plugging in suitable values for \(v\), we find \(C = D = 1\).
07

Integration

Now we can rewrite and integrate the expression: \(\frac{1}{4} \int \frac{v^2 x dx}{\left(v - 1\right)^{2}} = \frac{1}{4} \int \left(\frac{x dx}{v - 1} + \frac{x dx}{(v - 1)^2} \right)\) Integrating each term (notice that \(dv = 4x^3 dx\), so \(xDx=\frac{1}{4}d v\)): \(\frac{1}{16} \left(\ln |v - 1| - \frac{1}{v - 1} + K \right)\)
08

Reverse Substitution

Now we substitute back to get the result in terms of \(x\): \(\frac{1}{16} \left(\ln |x^4 - 1| - \frac{1}{x^4 - 1} + K \right)\) The final result of the integral (ii) is: \(\frac{1}{16} \left(\ln \left|\frac{x^4 - 1}{x^4 - 1} \right| + K \right)\)

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Most popular questions from this chapter

Evaluate the following integrals: (i) \(\int \frac{\sqrt{x^{4}+x^{-4}+2}}{x^{3}} d x\) (ii) \(\int \frac{d x}{\sqrt{2 x+3}+\sqrt{2 x-3}} d x\) (iii) \(\int \frac{(\sqrt{x}+1)\left(x^{2}-\sqrt{x}\right)}{x \sqrt{x}+x+\sqrt{x}} d x\) (iv) \(\int\left(\frac{1-x^{-2}}{x^{1 / 2}-x^{-1 / 2}}-\frac{2}{x^{3 / 2}}+\frac{x^{-2}-x}{x^{1 / 2}-x^{-1 / 2}}\right) d x\)

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