What is the integration of $\int \frac{(x {(tan x)}^{- 1}) d x}{{(1 + x^{2})}^{\frac{3}{2}}}$ $int((x(tanx)^-1)dx)/(1+x^2)^(3/2)$ ?

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What is the integration of $\int \frac{(x {(tan x)}^{- 1}) d x}{{(1 + x^{2})}^{\frac{3}{2}}}$ $int((x(tanx)^-1)dx)/(1+x^2)^(3/2)$ ?

$\int \frac{(x {(tan x)}^{- 1}) d x}{{(1 + x^{2})}^{\frac{3}{2}}}$ $int((x(tanx)^-1)dx)/(1+x^2)^(3/2)$

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Answer 1 · 2018-07-30T12:26:01

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

Explanation:

$\int \frac{(x {(tan x)}^{- 1}) d x}{{(1 + x^{2})}^{\frac{3}{2}}}$ $int((x(tanx)^-1)dx)/(1+x^2)^(3/2)$

Put $x = tan z$ $x=tanz$

$d x = {sec}^{2} z$ $dx=sec^2z$

$\int \frac{tan z . z . ({sec}^{2} z)}{{({sec}^{2} z)}^{\frac{3}{2}}} . d z$ $int (tanz.z.(sec^2z))/((sec^2z)^(3/2)).dz$

$\int \frac{tan z . z . d z}{sec z} . d z$ $int(tanz.z.dz)/secz.dz$

$\int (\frac{sin z}{cos z}) . cos z . z . d z$ $int(sinz/cosz).cosz.z.dz$

$\int sin z . z . d z$ $intsinz.z.dz$

Now I'm gonna use integration by parts

$\int u . d v = u . v - \int d u . v$ $intu.dv = u.v -intdu.v$

$u = z d z$ $u = z dz$
$d u = 1$ $du = 1$

$d v = sin z$ $dv=sinz$
$v = - cos z$ $v=-cosz$

Now put values in formula

$\int z . sin z . d z = z (- cos z) - \int (- cos z)$ $intz.sinz.dz=z(-cosz)-int(-cosz)$

$\int z . sin z . d z = z (- cos z) + \int cos z$ $intz.sinz.dz=z(-cosz)+intcosz$

$\int z . sin z . d z = z (- cos z) + sin z + C$ $intz.sinz.dz=z(-cosz)+sinz +C$ .......(1)

As we know $cos z = \frac{1}{sec z}$ $cosz=1/secz$

$sec z = \sqrt{1 + {tan}^{2} z}$ $secz=sqrt(1+tan^2z)$

And is equal to $tan z = x$ $tanz =x$ above.

then, $sec z = \sqrt{1 + x^{2}}$ $secz=sqrt(1+x^2)$

$cos z = \frac{1}{\sqrt{1 + x^{2}}}$ $cosz=1/sqrt(1+x^2)$

$sin z = \sqrt{1 - {cos}^{2} z}$ $sinz=sqrt(1-cos^2z)$

$sin z = \sqrt{1 - \frac{1}{1 + x^{2}}}$ $sinz=sqrt(1-1/(1+x^2))$

$sin z = \sqrt{\frac{x^{2}}{1 + x^{2}}}$ $sinz=sqrt(x^2/(1+x^2))$

$sin z = \frac{x}{\sqrt{1 + x^{2}}}$ $sinz = x/sqrt(1+x^2)$

Now put value of z, cosz, sinz in eq 1

Here is Final answer $\int \frac{(x {(tan x)}^{- 1}) d x}{{(1 + x^{2})}^{\frac{3}{2}}} = {tan}^{- 1} x (- \frac{1}{\sqrt{1 + x^{2}}}) + \frac{x}{\sqrt{1 + x^{2}}} + C$ $int((x(tanx)^-1)dx)/(1+x^2)^(3/2)=tan^-1x(-1/sqrt(1+x^2))+x/sqrt(1+x^2) +C$

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What is the integration of $\int \frac{(x {(tan x)}^{- 1}) d x}{{(1 + x^{2})}^{\frac{3}{2}}}$ $int((x(tanx)^-1)dx)/(1+x^2)^(3/2)$ ?

$\int \frac{(x {(tan x)}^{- 1}) d x}{{(1 + x^{2})}^{\frac{3}{2}}}$ $int((x(tanx)^-1)dx)/(1+x^2)^(3/2)$

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Explanation:

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What is the integration of int((x(tanx)^-1)dx)/(1+x^2)^(3/2)∫(x(tanx)−1)dx(1+x2)32int((x(tanx)^-1)dx)/(1+x^2)^(3/2) ?

int((x(tanx)^-1)dx)/(1+x^2)^(3/2)∫(x(tanx)−1)dx(1+x2)32int((x(tanx)^-1)dx)/(1+x^2)^(3/2)

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Explanation:

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What is the integration of $\int \frac{(x {(tan x)}^{- 1}) d x}{{(1 + x^{2})}^{\frac{3}{2}}}$ $int((x(tanx)^-1)dx)/(1+x^2)^(3/2)$ ?

$\int \frac{(x {(tan x)}^{- 1}) d x}{{(1 + x^{2})}^{\frac{3}{2}}}$ $int((x(tanx)^-1)dx)/(1+x^2)^(3/2)$