How do you evaluate the integral $\int ln (1 + x^{2})$ $int ln(1+x^2)$ ?

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How do you evaluate the integral $\int ln (1 + x^{2})$ $int ln(1+x^2)$ ?

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Answer 1 · 2017-01-20T03:39:54

$\int ln (1 + x^{2}) d x = x ln (1 + x^{2}) - 2 x + 2 arctan (x) + C$ $intln(1+x^2)dx=xln(1+x^2)-2x+2arctan(x)+C$

Explanation:

$I = \int ln (1 + x^{2}) d x$ $I=intln(1+x^2)dx$

Integration by parts takes the form $\int u d v = u v - \int v d u$ $intudv=uv-intvdu$ . For the given integral, let:

$u = ln (1 + x^{2})$ $u=ln(1+x^2)$

$u = ln (1 + x^{2}) d u = \frac{2 x}{1 + x^{2}} d x$ $color(white)(u=ln(1+x^2))du=(2x)/(1+x^2)dx$

$d v = d x$ $dv=dx$

$d v = d x v = x$ $color(white)(dv=dx)v=x$

Then:

$I = u v - \int v d u$ $I=uv-intvdu$

$I = x ln (1 + x^{2}) - \int \frac{2 x^{2}}{1 + x^{2}} d x$ $I=xln(1+x^2)-int(2x^2)/(1+x^2)dx$

$I = x ln (1 + x^{2}) - \int \frac{2 (1 + x^{2}) - 2}{1 + x^{2}} d x$ $I=xln(1+x^2)-int(2(1+x^2)-2)/(1+x^2)dx$

$I = x ln (1 + x^{2}) - \int (2 - \frac{2}{1 + x^{2}}) d x$ $I=xln(1+x^2)-int(2-2/(1+x^2))dx$

$I = x ln (1 + x^{2}) - 2 \int d x + 2 \int \frac{d x}{1 + x^{2}}$ $I=xln(1+x^2)-2intdx+2intdx/(1+x^2)$

$I = x ln (1 + x^{2}) - 2 x + 2 arctan (x) + C$ $I=xln(1+x^2)-2x+2arctan(x)+C$

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How do you evaluate the integral $\int ln (1 + x^{2})$ $int ln(1+x^2)$ ?

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Explanation:

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How do you evaluate the integral $\int ln (1 + x^{2})$ $int ln(1+x^2)$ ?