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∫1(2x+1)x2+2x+2 dx\int \frac{1}{(2x+1)\sqrt{x^{2}+2x+2}}\,dx

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Evaluate definite integrals using the chain ruleright chevron icon

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Q. ∫1(2x+1)x2+2x+2 dx\int \frac{1}{(2x+1)\sqrt{x^{2}+2x+2}}\,dx
  1. Identify Integral: Identify the integral to be solved.\newlineWe need to evaluate the integral of the function 1(2x+1)x2+2x+2\frac{1}{(2x+1)\sqrt{x^2+2x+2}} with respect to xx.
  2. Simplify Expression: Simplify the expression under the square root.\newlineNotice that x2+2x+1x^2 + 2x + 1 can be written as (x+1)2+1(x+1)^2 + 1, which is a sum of squares.
  3. Use Substitution: Use substitution to simplify the integral.\newlineLet u=x+1u = x + 1, which implies du=dxdu = dx. Then the integral becomes:\newline∫1(2(u−1)+1)u2+1du\int \frac{1}{(2(u-1)+1)\sqrt{u^2+1}}du
  4. Simplify Integral: Simplify the expression in the integral.\newlineThe integral now is:\newline∫1(2u−1)u2+1 du\int \frac{1}{(2u-1)\sqrt{u^2+1}}\,du
  5. Use Another Substitution: Use another substitution to transform the integral into a standard form.\newlineLet u2+1=v2u^2 + 1 = v^2, which implies 2u du=2v dv2u \, du = 2v \, dv. Then the integral becomes:\newline∫1(2v2−2)v2⋅(dvv)\int \frac{1}{(2v^2-2)\sqrt{v^2}} \cdot \left(\frac{dv}{v}\right)
  6. Simplify Expression: Simplify the expression in the integral.\newlineThe integral now is:\newline∫12v2−2 dv\int\frac{1}{2v^2-2} \, dv
  7. Factor Out Constant: Factor out the constant from the denominator.\newlineThe integral now is:\newline(\frac{\(1\)}{\(2\)})\int \frac{\(1\)}{v^\(2\)\(-1\)} \, dv
  8. Decompose Integrands: Decompose the integrand into partial fractions. The integral now is: \((\frac{1}{2})\int \frac{1}{(v-1)(v+1)} dv This can be decomposed into Av−1+Bv+1\frac{A}{v-1} + \frac{B}{v+1}.
  9. Find Constants: Find the constants AA and BB. By setting up the equation: 1=A(v+1)+B(v−1)1 = A(v+1) + B(v-1) and solving for AA and BB when v=1v = 1 and v=−1v = -1, we get A=12A = \frac{1}{2} and B=−12B = -\frac{1}{2}.
  10. Write with Partial Fractions: Write the integral with the partial fractions.\newlineThe integral now is:\newline(\frac{\(1\)}{\(2\)})\int(\frac{\(1\)}{\(2\)})/(v\(-1) - (\frac{11}{22})/(v+11)) \, dv
  11. Integrate Term by Term: Integrate term by term.\newlineThe integral now is:\newline(14)ln⁡∣v−1∣−(14)ln⁡∣v+1∣+C(\frac{1}{4})\ln|v-1| - (\frac{1}{4})\ln|v+1| + C
  12. Substitute Back for v: Substitute back for v and then for u.\newlineSubstituting back for v, we get:\newline(14)ln⁡∣u2−1∣−(14)ln⁡∣u2+1∣+C(\frac{1}{4})\ln|u^2-1| - (\frac{1}{4})\ln|u^2+1| + C\newlineAnd then substituting back for u, we get:\newline(14)ln⁡∣(x+1)2−1∣−(14)ln⁡∣(x+1)2+1∣+C(\frac{1}{4})\ln|(x+1)^2-1| - (\frac{1}{4})\ln|(x+1)^2+1| + C
  13. Substitute Back for u: Simplify the final expression.\newlineThe final answer is:\newline(\frac{\(1\)}{\(4\)})\ln|x^\(2+22x| - (\frac{11}{44})\ln|x^22+22x+22| + C

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