(1+x)/(3+x)(2+3x) dx
Answers
EXPLANATION.
∫(1 + x)/(3 + x)(2 + 3x)dx.
As we know that,
Partial fraction is applied only when coefficient of denominator > coefficient of numerator.
⇒ (1 + x)/(3 + x)(2 + 3x) = A/(3 + x) + B/(2 + 3x).
⇒ 1 + x = A(2 + 3x) + B(3 + x).
Put the value of x = -3 in equation, we get.
⇒ 1 + (-3) = A(2 + 3(-3)) + B(3 + (-3)).
⇒ 1 - 3 = A(2 - 9) + 0.
⇒ -2 = A(-7).
⇒ 2 = 7A.
⇒ A = 2/7.
Put the value of x = -2/3 in equation, we get.
⇒ 1 + (-2/3) = A[(2 + 3(-2/3)] + B(3 + (-2/3)].
⇒ 1 - 2/3 = 0 + B(3 - 2/3).
⇒ 3 - 2/3 = B(9 - 2/3).
⇒ 1/3 = B(7/3).
⇒ 1 = 7B.
⇒ B = 1/7.
Put the values in the equation, we get.
⇒ ∫(1 + x)/(3 + x)(2 + 3x)dx = ∫A/(3 + x)dx + ∫B/(2 + 3x)dx.
⇒ ∫2/7/(3 + x)dx + ∫1/7/(2 + 3x)dx.
⇒ ∫2/7(3 + x)dx + ∫dx/7(2 + 3x).
⇒ 2/7 ∫dx/(3 + x) + 1/7 ∫dx/(2 + 3x).
⇒ 2/7㏑(3 + x) + 1/7㏑(2 + 3x) + c.
MORE INFORMATION.
Integration of irrational functions.
(1) = If integral is in the form of = ∫dx/√ax² + bx + c, ∫√ax² + bx + c dx.
Then we integrate it by expressing ax² + bx + c = (x + α)² + β.
(2) = If the integrals of the form ∫px + q/√ax² + bx + c dx, ∫(px + q)√ax² + bx + c dx. first we express px + q in the form px + q = A[d(ax² + bx + c/dx] + B and then proceed as usual with standard form.