The correct option is C x 10ln|x+10|+C ∫xx+10dx =∫x+10 10x+10dx =∫x+10x+10dx 10∫dxx+10 =∫dx 10∫dxx+10 =x 10ln|x+10|+C ...

You visited us 2 times! Enjoying our articles? Unlock Full Access!

Higher Order Derivatives

Integrate ∫...

Question

Integrate $\int \frac{x}{x + 10} d x$

x + 10 ln | x + 10 | + C

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

x - 10 ln | x + 10 | + C

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

x + 10 ln | x | + C

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

None

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

Open in App

Solution

The correct option is C $x - 10 ln | x + 10 | + C$
$\int \frac{x}{x + 10} d x$

$= \int \frac{x + 10 - 10}{x + 10} d x$

$= \int \frac{x + 10}{x + 10} d x - 10 \int \frac{d x}{x + 10}$

$= \int d x - 10 \int \frac{d x}{x + 10}$

$= x - 10 ln | x + 10 | + C$

Suggest Corrections

Similar questions

I_{n} = \int_{0}^{1} \frac{(l n x)^{n}}{\sqrt{x}} d x, n ϵ N,

The value of the integral $\int (x^{2} + x) (x^{- 8} + 2 x^{- 9})^{1 / 10} d x$ is

\int \frac{10 x - 5}{\sqrt{x^{2} - x + 3} d x}

Integrate the following w.r.t $x : \frac{1}{2 x + 3}$

\frac{1}{1 + {(\log_{e} x)}^{2}}

\frac{\tan^{- 1} (\log_{e} x)}{x} + C

\tan^{- 1} (\log_{e} x) + C

\frac{\tan^{- 1} x}{x} + C

Join BYJU'S Learning Program