TS EAMCET · Maths · Indefinite Integration
- A
- B
- C
- D
Answer & Solution
Correct Answer
(D)
Step-by-step Solution
Detailed explanation
We have, I=∫x2-1x32x4-2x2+1dx ⇒I=∫x2-1x52-2x-2+x-4dx ⇒I=∫x-3-x-52-2x-2+x-4dx Put, 2-2x-2+x-4=t⇒2-2x-2+x-4=t2⇒4x-3-4x-5dx=2tdt ⇒I=12∫tdtt ⇒I=t2+c, where c is the constant of integration. ⇒I=122-2x2+1x4+c
See the Complete Solution
Get step-by-step explanations for this and 2.5 Lakh+ more JEE, NEET & CET questions.
- Unlock all solutions
- Practice the full chapter
- Track accuracy across PYQs
4.8 rated on Google Play · 14,000+ reviews
More questions from Maths
- The vertices of a triangle are \(A(1,7), B(-5,-1)\) and \(C(-1,2)\). Then, the equation of a bisector of the \(\angle A B C\) isTS EAMCET 2018 Hard
- Let \(z=x+i y\) be a complex number, \(A=\{z /|z| \leq 2\} \text { and } B=\{z /(1-i) z+(1+i) \bar{z} \geq 4\}\) Then which one of the following options belongs to \(A \cap B\) ?TS EAMCET 2020 Easy
- If a poisson variate \(X\) satisfies \(P(X=2)\) \(=P(X=3)\), then \(P(X=5)=\)TS EAMCET 2019 Medium
- If \((p, q)\) is the centre of the circle which cuts the three circles \(x^2+y^2-2 x-4 y+4=0, x^2+y^2+2 x-4 y+1=0\) and \(x^2+y^2-4 x-2 y-11=0\) orthogonally, then \(p+q=\)TS EAMCET 2024 Easy
- If the vectors \(\overline{B C}=2 \bar{i}+\bar{j}+\bar{k}\) and \(\overline{C D}=\bar{i}+2 \bar{j}-2 \bar{k}\) represent two adjacent sides of a parallelogram \(\mathrm{ABCD}\) and \(\theta\) is the angle between its diagonals \(\overline{A C}\) and \(\overline{B D}\) then \(\tan \theta=\)TS EAMCET 2023 Medium
- If \(u=f(r)\), where \(r^2=x^2+y^2\), then
\(\left(\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}\right)\) is equal toTS EAMCET 2012 Medium
More PYQs from TS EAMCET
- Let \(\mathrm{A}(\alpha, 4,7)\) and \(\mathrm{B}(3, \beta, 8)\) be two points in space. If YZ plane and ZX plane respectively divide the line segment joining the points A and B in the ratio \(2: 3\) and 4:5, then the point C which divides \(\overline{\mathrm{AB}}\) in the ratio \(\alpha: \beta\) externally isTS EAMCET 2025 Medium
- \(\mathrm{AU}^{235}\) nuclear reactor generates energy at a rate of \(3.70 \times 10^7 \mathrm{~J} / \mathrm{s}\). Each fission liberates \(185 \mathrm{MeV}\) useful energy. If the reactor has to operate for \(144 \times 10^4 \mathrm{~s}\), then, the mass of the fuel needed is (Assume Avogadro's number \(\left.=6 \times 10^{23} \mathrm{~mol}^{-1}, 1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J}\right)\)TS EAMCET 2013 Medium
- Let \(P\) represent the point \((3,6)\) on the parabola \(y^2=12 x\). For the parabola \(y^2=12 x\), if \(l_1\) is the length of the normal chord drawn at \(P\) and \(l_2\) is the length of the focal chord drawn through \(P\), then \(\frac{l_1}{l_2}=\)TS EAMCET 2019 Easy
- Consider a steady flow of oil in a pipeline. The cross-sectional radius of the pipeline decreases gradually as \(r=r_0 e^{-\alpha x}\), where \(\alpha=\frac{1}{3} \mathrm{~m}^{-1}\) and \(x\) is the distance from the pipeline inlet. If \(R_1\) is the Reynold's number for a certain pipeline cross-section at a distance \(x_1\) metre from the inlet and \(R_2\) is for distance \(\left(x_1+3\right)\) metre, then the ratio \(\frac{R_1}{R_2}\) isTS EAMCET 2020 Medium
- Let \(\alpha \neq 1\) be a real root of the equation \(x^3-a x^2+a x-1=0\), where \(a \neq-1\) is a rea number. Then, a root of this equation, among the following, isTS EAMCET 2010 Medium
- \(\sin A+\sin B=\sqrt{3}(\cos B-\cos A)\) \(\Rightarrow \sin 3 A+\sin 3 B\) is equal toTS EAMCET 2007 Hard