Question

Evaluate the integral by first performing long division on the integrand and then writing the proper fraction as a sum of partial fractions. \int \frac{3x^3 + 3x^2 + 5}{x^2 + x} dx 5 \ln|x+1| - 5 \ln|x| + C \frac{3}{2}x^2 - 5 \ln|x+1| + 5 \ln|x| + C \frac{3}{2}x^2 + 5 \ln|x-1| - 5 \ln|x| + C 3x^2 + 5 \ln|x+1| - 5 \ln|x| + C

          Evaluate the integral by first performing long division on the integrand and then writing the proper fraction as a sum of partial fractions.
\int \frac{3x^3 + 3x^2 + 5}{x^2 + x} dx
5 \ln|x+1| - 5 \ln|x| + C
\frac{3}{2}x^2 - 5 \ln|x+1| + 5 \ln|x| + C
\frac{3}{2}x^2 + 5 \ln|x-1| - 5 \ln|x| + C
3x^2 + 5 \ln|x+1| - 5 \ln|x| + C

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$Evaluate the integral by first performing long division on the integrand and then writing the proper fraction as a sum of partial fractions. ∫(3x^3 + 3x^2 + 5)/(x^2 + x) dx 5 ln|x+1| - 5 ln|x| + C (3)/(2)x^2 - 5 ln|x+1| + 5 ln|x| + C (3)/(2)x^2 + 5 ln|x-1| - 5 ln|x| + C 3x^2 + 5 ln|x+1| - 5 ln|x| + C$

Added by Yvonne B.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Suman Saurav Thakur

10/19/2021

Step 1

Dividing \(3x^3 + 3x^2 + 5\) by \(x^2 + x\), we get: \(3x + 3 - \frac{3x + 3}{x+1}\) Show more…

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Evaluate the integral by first performing long division on the integrand and then writing the proper fraction as a sum of partial fractions. int (3x^(3)+3x^(2)+5)/(x^(2)+x)dx 5ln|x+1|-5ln|x|+C (3)/(2)x^(2)-5ln|x+1|+5ln|x|+C (3)/(2)x^(2)+5ln|x-1|-5ln|x|+C 3x^(2)+5ln|x+1|-5ln|x|+C Evaluate the integral by first performing long division on the integrand and then writing the proper fraction as a sum of partial fractions. x2+x O51n|x+1|-51n||+C O2-5 In|x+1|+5 In|x|+C O32+5In|x+1|-5In|x|+C

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