$\int \frac{n x^{n - 1} + m^{x} l o g m}{x^{n} + m^{x}} d x$ equal

m^{x} - x^{n} + C

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

m^{x} + x^{n} + C

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

(m^{x} - x^{n})^{- 1} + C

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

l o g (m^{x} + x^{n}) + C

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

Open in App

Solution

The correct option is D $l o g (m^{x} + x^{n}) + C$
$\int \frac{n x^{n - 1} + m^{x} log m}{x^{n} + m^{x}} d x$

substitute $u = x^{n} + m^{x} \to d u = \frac{1}{n x^{n - 1} + m^{x} log m} d x$

$= \int \frac{1}{u} d u$

$= log u$

$= log (| x^{n} + m^{x} |) + C$

Suggest Corrections

Similar questions

I = \int \frac{e^{x} (1 + n x^{n - 1} - x^{2 n})}{(1 - x^{n} \sqrt{1 - x^{2 n}})} d x = e^{x} \sqrt{\frac{1 + x^{n}}{1 - x^{n}}}

x_{1}, x_{2}, x_{3}, x_{4} . . . . x_{2 n + 1}

[\frac{(x_{2 n + 1} - x_{1})}{(x_{2 n + 1} + x_{1})}] + [\frac{(x_{2 n} - x_{2})}{(x_{2 n} + x_{2})}] . . . . . . . . . . . . . . . . . . . . [\frac{(x_{n + 2 - x_{n}})}{(x_{n + 2} + x_{n})}]

\frac{x^{n} - y^{n}}{x^{2} - y^{2}}

x^{2} \neq y^{2}

n

6

\int e^{x} (\frac{1 + n x^{n - 1} - x^{2 n}}{(1 - x^{n}) \sqrt{1 - x^{2 n}}}) d x

x^{n}

n x^{n - 1} .

Join BYJU'S Learning Program