The correct option is B x log (1+x^2) 2x+2 tan^ 1x + c ∫ log(1+x^2)dx=∫ log(1+x^2)1dx =xlog(1+x^2) ∫11+x^2x #160;dx =xlog(1+x^2) 2∫ [x^2 ...

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Functions

Solve: ∫log1...

Question

Solve:
$\int log (1 + x^{2}) d x$

x l o g (1 + x^{2}) - 2 x + 2 {tan}^{- 1} x + c

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x l o g (1 + x^{2}) - 2 x - 2 {tan}^{- 1} x + c

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l o g (1 + x^{2}) - 2 x + 2 {tan}^{- 1} x + c

No worries! We‘ve got your back. Try BYJU‘S free classes today!

l o g (1 + x^{2}) + 2 x - 2 {tan}^{- 1} x + c

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Solution

The correct option is B $x l o g (1 + x^{2}) - 2 x + 2 {tan}^{- 1} x + c$
$\int l o g (1 + x^{2}) d x = \int l o g (1 + x^{2}) 1 d x$

$= x l o g (1 + x^{2}) - \int \frac{1}{1 + x^{2}} x d x$

$= x l o g (1 + x^{2}) - 2 \int [\frac{x^{2} + 1}{(x^{2} + 1)} - \frac{1}{(x^{2} + 1)}] d x$

$= x l o g (1 + x^{2}) - 2 [\int 1 - d x - \int \frac{1}{(x^{2} + 1)} d x]$

$= x l o g (1 + x^{2}) - 2 x + 2 t a n^{- 1} x + c$

$\int l o g (1 + x^{2}) d x = x l o g (1 + x^{2}) - 2 x + 2 t a n^{- 1} x + c$

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