_ ^ 10 x^9 + 10 ^x _e 10 10 ^x + x^ 10 ;dx = - Vidyadip

MCQ

$\int_{}^{} {\frac{{10{x^9} + {{10}^x}{{\log }_e}10}}{{{{10}^x} + {x^{10}}}}} \;dx = $

A
$ - \frac{1}{2}\frac{1}{{{{({{10}^x} + {x^{10}})}^2}}} + c$
✓
$\log ({10^x} + {x^{10}}) + c$
C
$\frac{1}{2}\frac{1}{{{{({{10}^x} + {x^{10}})}^2}}} + c$
D
None of these

✓

Answer

Correct option: B.

$\log ({10^x} + {x^{10}}) + c$

b
(b) Put ${x^{10}} + {10^x} = t \Rightarrow (10{x^9} + {10^x}{\log _e}10)\,dx = dt,$
then $\int_{}^{} {\frac{{10{x^9} + {{10}^x}{{\log }_e}10}}{{{{10}^x} + {x^{10}}}}\,dx} = \int_{}^{} {\frac{1}{t}\,dt = \log t + c} $
$ = \log ({x^{10}} + {10^x}) + c.$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from STD 12 - 7.1 inderfinite integral→STD 12 - 7.1 inderfinite integral — all question types→Mathematics STD 12 Science question papers→CBSE Board STD 12 Science question papers→CBSE Board question paper generator→

Similar questions

10 persons are seated around a round table. What is the probability that 4 particular persons are always seated together?

$\int_0^{2n\pi } {\left( {|\sin x| - \left. {\left| {\frac{1}{2}\sin x} \right.} \right|} \right)} \;dx$ equals

Let the relation $R$ in the set $A=\{x \in Z: 0 \leq x \leq 12\}$, given by $R=\{(a, b):|a-b|$ is a multiple of 4. $\}$ Then [1], the equivalence class containing 1 , is

$\sin\Big\{2\cos^{-1}\Big(\frac{-3}{5}\Big)\Big\}$ is equal to:

If $y = \sec {x^0}$, then ${{dy} \over {dx}} = $

The area enclosed by the curve $\frac{\text{x}^2}{25}+\frac{\text{y}^2}{9}=1$ is:

Let $\quad \overrightarrow{ a }=\alpha \hat{ i }+3 \hat{ j }-\hat{ k }, \overrightarrow{ b }=3 \hat{ i }-\beta \hat{ j }+4 \hat{ k } \quad$ and $\overrightarrow{ c }=\hat{ i }+2 \hat{ j }-2 \hat{ k }$ where $\alpha, \beta \in R$, be three vectors. If the projection of $\vec{a}$ on $\vec{c}$ is $\frac{10}{3}$ and $\overrightarrow{ b } \times \overrightarrow{ c }=-6 \hat{ i }+10 \hat{ j }+7 \hat{ k }$, then the value of $\alpha+\beta$ equal to

Let $\vec{u}=2 \hat{i}-\hat{j}+\hat{k}, \vec{v}=-3 \hat{j}+2 \hat{k}$ be vectors in $R^3$ and $\vec{w}$ be a unit vector in the $X Y$-plane. Then, the

The lines $\overrightarrow{ r }=(\hat{ i }-\hat{ j })+\ell(2 \hat{ i }+\hat{ k })$ and $\overrightarrow{ r }=(2 \hat{ i }-\hat{ j })+ m (\hat{ i }+\hat{ j }-\hat{ k })$

If a relation $R$ is defined on $A=\{1,2,3\}$, then $R$ is, where $R =\{(1,1),(2,2),(3,3),(1,2),(2,3)$, $(1,3)\}$ :