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Partial Fraction Decomposition Calculator

Find partial fractions step by step

This online calculator will find the partial fraction decomposition of the rational function, with steps shown.

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Solution

Your input: perform the partial fraction decomposition of t(1−t)2(t+1)(−t2−2t+1)

Simplify the expression: t(1−t)2(t+1)(−t2−2t+1)=−t(t−1)2(t+1)(t2+2t−1)

Factor the denominator: −t(t−1)2(t+1)(t2+2t−1)=−t(t−1)2(t+1)(t+1+√2)(t−√2+1)

The form of the partial fraction decomposition is

−t(t−1)2(t+1)(t+1+√2)(t−√2+1)=At−1+B(t−1)2+Ct+1+Dt+1+√2+Et−√2+1

Write the right-hand side as a single fraction:

−t(t−1)2(t+1)(t+1+√2)(t−√2+1)=(t−1)2(t+1)(t+1+√2)E+(t−1)2(t+1)(t−√2+1)D+(t−1)2(t+1+√2)(t−√2+1)C+(t−1)(t+1)(t+1+√2)(t−√2+1)A+(t+1)(t+1+√2)(t−√2+1)B(t−1)2(t+1)(t+1+√2)(t−√2+1)

The denominators are equal, so we require the equality of the numerators:

−t=(t−1)2(t+1)(t+1+√2)E+(t−1)2(t+1)(t−√2+1)D+(t−1)2(t+1+√2)(t−√2+1)C+(t−1)(t+1)(t+1+√2)(t−√2+1)A+(t+1)(t+1+√2)(t−√2+1)B

Expand the right-hand side:

−t=t4A+t4C+t4D+t4E+2t3A+t3B−√2t3D+√2t3E−2t2A+3t2B−4t2C−2t2D+√2t2D−2t2E−√2t2E−2tA+tB+4tC+√2tD−√2tE+A−B−C−√2D+D+E+√2E

Collect up the like terms:

−t=t4(A+C+D+E)+t3(2A+B−√2D+√2E)+t2(−2A+3B−4C−2D+√2D−2E−√2E)+t(−2A+B+4C+√2D−√2E)+A−B−C−√2D+D+E+√2E

The coefficients near the like terms should be equal, so the following system is obtained:

{A+C+D+E=02A+B−√2D+√2E=0−2A+3B−4C−2D+√2D−2E−√2E=0−2A+B+4C+√2D−√2E=−1A−B−C−√2D+D+E+√2E=0

Solving it (for steps, see system of equations calculator), we get that A=38, B=−14, C=−18, D=−18+√28, E=−√28−18

Therefore,

−t(t−1)2(t+1)(t+1+√2)(t−√2+1)=38t−1+−14(t−1)2+−18t+1+−18+√28t+1+√2+−√28−18t−√2+1

Answer: t(1−t)2(t+1)(−t2−2t+1)=38t−1+−14(t−1)2+−18t+1+−18+√28t+1+√2+−√28−18t−√2+1