Evaluate - 10 x 9 - 10 x log 10 - x 10 - 10 x dx

Let x^10+10^x=t ⇒(10x^9+10^xlog10)dx=dt ∫10x^9+10^xlog10x^10+10^xdx =∫dtt =log|t|+c =log|x^10+10^x|+c where t=x^10+10^x ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Integration by Substitution

Evaluate ∫ ...

Question

Evaluate $\int \frac{10 x^{9} + 10^{x} log 10}{x^{10} + 10^{x}} d x$

Open in App

Solution

Let $x^{10} + 10^{x} = t$
$\Rightarrow (10 x^{9} + 10^{x} log 10) d x = d t$
$\int \frac{10 x^{9} + 10^{x} log 10}{x^{10} + 10^{x}} d x$
$= \int \frac{d t}{t}$
$= log | t | + c$
$= log ∣ ∣ x^{10} + 10^{x} ∣ ∣ + c$ where $t = x^{10} + 10^{x}$

Suggest Corrections

Similar questions

\int \frac{10 x^{9} + 10^{x} {log}_{e} 10}{x^{10} + 10^{x}} d x =

Choose the correct. answer.
$\int (\frac{10 x^{9} + 10^{x} l o g_{e} 10}{x^{10} + 10^{x}}) d x$ equals
$(a) 10^{x} + x^{10} + C (b) 10^{x} + x^{10} + C (c) (10^{x} - x^{10})^{- 1} + C (d) l o g | 10^{x} + x^{10} | + C$

\int \frac{10 x^{9} + 10^{x} {log}_{e} 10 d x}{x^{10} + 10^{x}}

equals

\int \frac{10 x^{9} + 10^{x} {log}_{e^{10}} d x}{x^{10} + 10^{x}}

equals

Q. Evaluate:

\int \frac{2^{x + 1} - 5^{x - 1}}{10^{x}} d x

Join BYJU'S Learning Program

Evaluate ∫10x9+10xlog10x10+10xdx

Let x10+10x=t⇒(10x9+10xlog10)dx=dt∫10x9+10xlog10x10+10xdx=∫dtt=log|t|+c=log∣∣x10+10x∣∣+c where t=x10+10x

Evaluate $\int \frac{10 x^{9} + 10^{x} log 10}{x^{10} + 10^{x}} d x$

Let $x^{10} + 10^{x} = t$
$\Rightarrow (10 x^{9} + 10^{x} log 10) d x = d t$
$\int \frac{10 x^{9} + 10^{x} log 10}{x^{10} + 10^{x}} d x$
$= \int \frac{d t}{t}$
$= log | t | + c$
$= log ∣ ∣ x^{10} + 10^{x} ∣ ∣ + c$ where $t = x^{10} + 10^{x}$