Question: Is it possible to get multiple correct results when evaluating an indefinite integral? If I use two different techniques to evaluate an integral, and I get two different answers, have I necessarily done something wrong?
Often, an indefinite integral can be evaluated using different techniques. For example, an integrand might be simplified via partial fractions or other algebraic techniques before integration, or it might be amenable to a clever substitution. These techniques give different results. For example, looking over a few other questions on MSE:
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From this question: evaluate $$ \int x(x^2+2)^4\,\mathrm{d}x. $$
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Via the substitution $u = x^2+2$, this becomes $$ \int x(x^2+2)^4\,\mathrm{d}x = \frac{1}{10}x^{10} + x^8 + 4x^6 + 8x^4 + 8x^2 + \frac{32}{5} + C. $$
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However, multiplying out the polynomial and integrating using the power rule gives $$ \int x(x^2+2)^4\,\mathrm{d}x = \frac{1}{10}x^{10} + x^8 + 4x^6 + 8x^4 + 8x^2 + C $$
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From this question: evaluate $$ \int \frac{1-x}{(x+1)^2} \,\mathrm{d}x. $$
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Simplifying the integrand using partial fractions then integrating gives $$ \int \frac{1-x}{(x+1)^2} \,\mathrm{d}x = \frac{2}{(x+1)} - \ln|x+1| + C. $$
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Via integration by parts, we get $$ \int \frac{1-x}{(x+1)^2} \,\mathrm{d}x = \frac{x-1}{(x+1)} - \ln|x+1| + C. $$
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From this question: evaluate $$ \int \frac {\tan(\pi x)\sec^2(\pi x)}2\,\mathrm{d}x. $$
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Using the substitution $u = \sec(\pi x)$, this becomes $$ \int \frac {\tan(\pi x)\sec^2(\pi x)}2\,\mathrm{d}x = \frac {\sec^2(\pi x)}{4\pi} + C. $$
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Using the substitution $u = \tan(\pi x)$, this becomes $$ \int \frac {\tan(\pi x)\sec^2(\pi x)}2\,\mathrm{d}x = \frac {\tan^2(\pi x)}{4\pi} + C. $$
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from Hot Weekly Questions - Mathematics Stack Exchange
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