Calculus

Integrals

Medium

common

$\int_{4}^{10} \frac{\log(x^2)}{\log(x^2)+\log(196-28x+x^2)}dx$ is equal to : (Consider $\log_e x = \log x$ )

✅ Correct Option: 3

Note $196-28x+x^2=(14-x)^2$ , so $f(x)=\frac{\log x^2}{\log x^2+\log(14-x)^2}$ .

Since $4+10=14$ , replacing $x$ by $14-x$ swaps the two logarithms, giving $f(x)+f(14-x)=1$ .

Hence $2I=\int_4^{10}1\,dx=6$ , so $I=3$ .

2026: 14 May Shift 2

Match the LIST-I with LIST-II

	LIST-I (Definite integrals)		LIST-II (Value)
A.	$\displaystyle\int_{-6}^{6} \vert x \vert \, dx$	I.	$1$
B.	$\displaystyle\int_{-\pi/2}^{\pi/2} \sin^{2017} x \, dx$	II.	$36$
C.	$\displaystyle\int_{0}^{\pi/4} \cos x \, dx$	III.	$0$
D.	$\displaystyle\int_{0}^{2} \dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{2-x}} \, dx$	IV.	$\dfrac{1}{\sqrt{2}}$

Choose the correct answer from the options given below:

2025: 22 May Shift 1

The Value of $\int_1^3 |2x - 1|dx$ equal to

2025: 15 May Shift 2

For $x > 1$ , $\int \frac{e^{7\log x} - e^{5\log x}}{e^{5\log x} - e^{4\log x}} dx$ equals.

2025: 2 June Shift 1

The value of $\int_{-a}^a f(x)dx$ where $f(x) = \frac{7^x}{1+7^x}$ is:

2022: 30 Aug Shift 1

$\int x\sqrt{x + 2} \, dx$ is equal to :

2026: 22 May Shift 1

Value of $\int_{-2}^{2}|2x+1|dx$ is:

∫410log⁡(x2)log⁡(x2)+log⁡(196−28x+x2)dx\int_{4}^{10} \frac{\log(x^2)}{\log(x^2)+\log(196-28x+x^2)}dx∫410​log(x2)+log(196−28x+x2)log(x2)​dx is equal to : (Consider log⁡ex=log⁡x\log_e x = \log xloge​x=logx)