MHT CET · Maths · Indefinite Integration
The value of \(\mathrm{I}=\int \frac{\mathrm{d} x}{x^2\left(x^4+1\right)^{\frac{3}{4}}}\) is
- A \(-\left(x^4+1\right)^{\frac{1}{4}}+\mathrm{c}\), (where c is a constant of integration)
- B \(\left(x^4+1\right)^{\frac{1}{4}}+\mathrm{c}\), (where c is a constant of integration)
- C \(\left(1+\frac{1}{x^4}\right)^{\frac{1}{4}}+\mathrm{c}\), (where c is a constant of integration)
- D \(-\left(1+\frac{1}{x^4}\right)^{\frac{1}{4}}+\mathrm{c}\), (where c is a constant of integration)
Answer & Solution
Correct Answer
(D) \(-\left(1+\frac{1}{x^4}\right)^{\frac{1}{4}}+\mathrm{c}\), (where c is a constant of integration)
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& \text { Let } \mathrm{I}=\int \frac{1}{x^2\left(x^4+1\right)^{\frac{3}{4}}} \mathrm{~d} x=\int \frac{\mathrm{d} x}{x^5\left(1+\frac{1}{x^4}\right)^{\frac{3}{4}}} \\
& \text { Put } 1+\frac{1}{x^4}=\mathrm{t} \Rightarrow \frac{-4}{x^5} \mathrm{~d} x=\mathrm{dt} \\
& \therefore \quad I=-\frac{1}{4} \int \frac{d t}{t^{\frac{3}{4}}}=-\frac{1}{4} \times 4 t^{\frac{1}{4}}+c=-t^{\frac{1}{4}}+c \\
& =-\left(1+\frac{1}{x^4}\right)^{\frac{1}{4}}+c
\end{aligned}\)
& \text { Let } \mathrm{I}=\int \frac{1}{x^2\left(x^4+1\right)^{\frac{3}{4}}} \mathrm{~d} x=\int \frac{\mathrm{d} x}{x^5\left(1+\frac{1}{x^4}\right)^{\frac{3}{4}}} \\
& \text { Put } 1+\frac{1}{x^4}=\mathrm{t} \Rightarrow \frac{-4}{x^5} \mathrm{~d} x=\mathrm{dt} \\
& \therefore \quad I=-\frac{1}{4} \int \frac{d t}{t^{\frac{3}{4}}}=-\frac{1}{4} \times 4 t^{\frac{1}{4}}+c=-t^{\frac{1}{4}}+c \\
& =-\left(1+\frac{1}{x^4}\right)^{\frac{1}{4}}+c
\end{aligned}\)
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
- If \(\mathrm{e}^{\mathrm{y}}+x \mathrm{y}=\mathrm{e}\), then the ordered pair \(\left(\frac{\mathrm{dy}}{\mathrm{d} x}, \frac{\mathrm{~d}^2 \mathrm{y}}{\mathrm{d} x^2}\right)\) at \(x=0\) is equal toMHT CET 2025 Medium
- The general solution of the differential equation \(\left(1+y^{2}\right)+\left(x-e^{\tan ^{-1} y}\right) \frac{d y}{d x}=0\) isMHT CET 2020 Medium
- The equation of pair of lines \(y=\mathrm{p} x\) and \(y=\mathrm{q} x\) can be written as \((y-\mathrm{p} x)(y-\mathrm{q} x)=0\). Then the equation of the pair of the angle bisectors of the lines \(x^2-4 x y-5 y^2=0\) isMHT CET 2024 Hard
- If \(A\) and \(B\) are the foot of the perpendicular drawn from the point \(\mathrm{Q}(\mathrm{a}, \mathrm{b}, \mathrm{c})\) to the planes \(\mathrm{YZ}\) and \(\mathrm{ZX}\) respectively, then the equation of the plane through the points \(\mathrm{A}, \mathrm{B}\), and \(\mathrm{O}\) is (where \(\mathrm{O}\) is the origin)MHT CET 2021 Medium
- \(\int_0^\pi \frac{\mathrm{d} x}{4+3 \cos x}=\)MHT CET 2023 Hard
- The angle between the lines \(\bar{r}=(3 \hat{i}+2 \hat{j}-4 \hat{k})+\lambda(\hat{i}+2 \hat{j}+2 \hat{k})\) and \(\bar{r}=(5 \hat{i}-2 \widehat{k})+\mu(3 \hat{i}+2 \hat{j}+6 \widehat{k})\)MHT CET 2022 Easy
More PYQs from MHT CET
- What is the volume of 1 mole of a crystalline solid having unit cell edge length \(16 \times 10^{-8} \mathrm{~cm}\), if it's unit cell contains 24 molecules?MHT CET 2020 Medium
- What will happen to the developing foetus if corpus luteum regresses in third week of pregnancy?MHT CET 2022 Hard
- The resultant of two vectors \(\vec{A}\) and \(\vec{B}\) is \(\vec{C}\). If the magnitude of \(\vec{B}\) is doubled, the new resultant vector becomes perpendicular to \(\overrightarrow{\mathrm{A}}\). Then the magnitude of \(\overrightarrow{\mathrm{C}}\) isMHT CET 2020 Medium
- Which of the following amines gives yellow oily liquid with \(\mathrm{HNO}_{2}\) ?MHT CET 2010 Hard
- A physical quantity \(A\) can be determined by measuring parameters \(B\), \(C, D\) and \(E\) using the relation \(A=\frac{B^\alpha C^\beta}{D^\gamma E^\delta}\). If the maximum errors in the measurement are b %, c %, d % and e % then maximum error in the value of \(A\) isMHT CET 2025 Easy
- The equation of the line passing through \((1,2,3)\) and perpendicular to the lines
\(x-1=\frac{y+2}{2}=\frac{z+4}{4}\) and \(\frac{x-1}{2}=\frac{y-2}{2}=z+3\) isMHT CET 2020 Easy