Integrate. What is the next step ?

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Integrate. What is the next step ?

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Answer 1 · 2018-04-12T13:58:10

$\frac{x^{3}}{3} + \frac{35}{12} ln | x - 2 | + \frac{61}{24} ln (x^{2} + 2 x + 4) + \frac{1}{4 \sqrt{3}} {tan}^{- 1} (\frac{x + 1}{\sqrt{3}}) + C$ $x^3/3+35/12 ln|x-2|+61/24 ln(x^2+2x+4)+1/(4sqrt3)tan^-1((x+1)/sqrt3)+C$

Explanation:

You have done almost all the heavy lifting here! The first of the two integrals is really simple,

$\frac{35}{12} \int \frac{1}{x - 2} d x = \frac{35}{12} ln | x - 2 |$ $35/12int 1/(x-2) dx = 35/12 ln|x-2|$

so I am guessing you need help with the second:

$\frac{1}{12} \int \frac{61 x + 64}{x^{2} + 2 x + 4} d x$ $1/12 int (61x+64)/(x^2+2x+4) dx$

For this, note that

$d (x^{2} + 2 x + 4) = (2 x + 2) d x = 2 (x + 1) d x$ $d(x^2+2x+4)=(2x+2)dx=2(x+1)dx$

so that it makes sense to split up the numerator into

$61 x + 64 = 61 (x + 1) + 3$ $61x+64=61(x+1)+3$

Then

$\frac{1}{12} \int \frac{61 x + 64}{x^{2} + 2 x + 4} d x = \frac{61}{12} \int \frac{x + 1}{x^{2} + 2 x + 4} d x + \frac{1}{12} \int \frac{3}{x^{2} + 2 x + 4} d x$ $1/12 int (61x+64)/(x^2+2x+4)dx=61/12 int (x+1)/(x^2+2x+4)dx +1/12 int 3/(x^2+2x+4)dx$

The first of these is

$\frac{61}{24} \int \frac{2 (x + 1) d x}{x^{2} + 2 x + 4} = \frac{61}{24} ln (x^{2} + 2 x + 4)$ $61/24 int (2(x+1)dx)/(x^2+2x+4) =61/24 ln(x^2+2x+4)$

while the second is

$\frac{1}{4} \int \frac{d x}{x^{2} + 2 x + 4} = \frac{1}{4} \int \frac{d x}{{(x + 1)}^{2} + 3} = \frac{1}{4 \sqrt{3}} {tan}^{- 1} (\frac{x + 1}{\sqrt{3}})$ $1/4 int (dx)/(x^2+2x+4) = 1/4 int dx/((x+1)^2+3) = 1/(4sqrt3) tan^-1((x+1)/sqrt3)$

Adding all the terms (including the $\int x^{2} d x$ $int x^2 dx$ ) we get the integral :

$\frac{x^{3}}{3} + \frac{35}{12} ln | x - 2 | + \frac{61}{24} ln (x^{2} + 2 x + 4) + \frac{1}{4 \sqrt{3}} {tan}^{- 1} (\frac{x + 1}{\sqrt{3}}) + C$ $x^3/3+35/12 ln|x-2|+61/24 ln(x^2+2x+4)+1/(4sqrt3)tan^-1((x+1)/sqrt3)+C$

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Integrate. What is the next step ?

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