Evaluate integration 1÷(x+1)(x+4)dx ?

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Evaluate integration 1÷(x+1)(x+4)dx ?

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Answer 1 · 2018-03-13T09:17:20

$I = \frac{1}{3} (ln (x + 1) - ln (x + 4)) + C$ $I=1/3(ln(x+1)-ln(x+4))+C$

Explanation:

We want to integrate

$I = \int \frac{1}{(x + 1) (x + 4)} d x$ $I=int1/((x+1)(x+4))dx$

By partial fraction decomposition

$I = \int \frac{A}{x + 1} + \frac{B}{x + 4} d x$ $I=intA/(x+1)+B/(x+4)dx$

Find the constants $A$ $A$ and $B$ $B$

$1 = A (x + 4) + B (x + 1)$ $1=A(x+4)+B(x+1)$

Let $x = - 4$ $x=-4$ then $B = - \frac{1}{3}$ $B=-1/3$
Let $x = - 1$ $x=-1$ then $A = \frac{1}{3}$ $A=1/3$

Thus

$I = \frac{1}{3} \int \frac{1}{x + 1} - \frac{1}{x + 4} d x$ $I=1/3int1/(x+1)-1/(x+4)dx$

$I = \frac{1}{3} (ln (x + 1) - ln (x + 4)) + C$ $color(white)(I)=1/3(ln(x+1)-ln(x+4))+C$

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Evaluate integration 1÷(x+1)(x+4)dx ?

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