How do you evaluate the integral $\int ln (x^{2} + x^{3})$ $int ln(x^2+x^3)$ ?

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How do you evaluate the integral $\int ln (x^{2} + x^{3})$ $int ln(x^2+x^3)$ ?

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Answer 1 · 2017-01-14T01:05:40

$x ln (x^{2} + x^{3}) + ln (| x - 1 |) - 3 x + C$ $xln(x^2+x^3)+ln(abs(x-1))-3x+C$

Explanation:

We have the integral:

$I = \int ln (x^{2} + x^{3}) d x$ $I=intln(x^2+x^3)dx$

We should apply integration by parts, which takes the form $\int u d v = u v - \int v d u$ $intudv=uv-intvdu$ . Where $\int ln (x^{2} + x^{3}) d x = \int u d v$ $intln(x^2+x^3)dx=intudv$ , let $u = ln (x^{2} + x^{3})$ $u=ln(x^2+x^3)$ and $d v = d x$ $dv=dx$ .

Differentiating $u$ $u$ and integrating $d v$ $dv$ gives:

${(u=ln(x^2+x^3)" "=>" "du=(2x+3x^2)/(x^2+x^3)dx=(2+3x)/(x+x^2)dx),(dv=dx" "=>" "v=x):}$

So:

$I=uv-intvdu$

$I=xln(x^2+x^3)-int(2x+3x^2)/(x+x^2)dx$

$I=xln(x^2+x^3)-int(2+3x)/(1+x)dx$

Either perform long division on $(3x+2)/(x+1)$ or do the following rewriting: $(3x+2)/(x+1)=(3(x+1)-1)/(x+1)=3-1/(x-1)$ . So:

$I=xln(x^2+x^3)-(3intdx-intdx/(x-1))$

$I=xln(x^2+x^3)-(3x-ln(abs(x-1)))$

$I=xln(x^2+x^3)+ln(abs(x-1))-3x+C$

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How do you evaluate the integral $\int ln (x^{2} + x^{3})$ $int ln(x^2+x^3)$ ?

1 Answer

Explanation:

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How do you evaluate the integral $\int ln (x^{2} + x^{3})$ $int ln(x^2+x^3)$ ?