Qus : 1 2
$\int f(x)\mathrm{d}x=g(x)$, then $\int {x}^5f({x}^3)\mathrm{d}x$
1 $$\frac{1}{3}{x}^3g({x}^3)-3\int {x}^4g({x}^3)\mathrm{d}x+C$$ 2 $$\frac{1}{3}{x}^3g({x}^3)-\int {x}^2g({x}^3)\mathrm{d}x+C$$ 3 $$\frac{1}{3}{x}^3g({x}^3)-\int {x}^3g({x}^3)\mathrm{d}x+C$$ 4 None of these Go to Discussion Solution
Quick Solution
Given:
\( \int f(x)\, dx = g(x) \)
Required: \( \int x^5 f(x^3)\, dx \)
Use substitution:
Let \( u = x^3 \Rightarrow du = 3x^2\, dx \Rightarrow dx = \frac{du}{3x^2} \)
Now rewrite the integral:
\[
\int x^5 f(x^3)\, dx
= \int x^5 f(u) \cdot \frac{du}{3x^2}
= \frac{1}{3} \int x^3 f(u)\, du
\]
But \( x^3 = u \), so:
\[
\frac{1}{3} \int u f(u)\, du
\]
Now integrate by parts or use the identity:
\[
\int u f(u)\, du = u g(u) - \int g(u)\, du
\]
Final answer:
\[
\int x^5 f(x^3)\, dx = \frac{1}{3} \left[ x^3 g(x^3) - \int g(x^3) \cdot 3x^2\, dx \right]
= x^3 g(x^3) - \int x^2 g(x^3)\, dx
\]
\[
\boxed{ \int x^5 f(x^3)\, dx = x^3 g(x^3) - \int x^2 g(x^3)\, dx }
\]
Qus : 2 3
If $\int x\, \sin x\, sec^3x\, dx=\frac{1}{2}\Bigg{[}f(x){se}c^2x+g(x)\Bigg{(}\frac{\tan x}{x}\Bigg{)}\Bigg{]}+C$, then which of the following is true?
1 $$f(x)-g(x)=0$$ 2 $$f(x).g(x)=0$$ 3 $$f(x)+g(x)=0$$ 4 $$f(x)+g(x)=1$$ Go to Discussion
Solution Qus : 4 2 If
, then the values of A
1 , A
2 , A
3 , A
4 are
1 A1
= 1/2 , A2 =1/4 , A3 = 1/6 , A4 = 1/8 2 A1
= 1/8, A2 =1/16 , A3 = 1/24 , A4 = 1/32 3 A1
= 1/6, A2 =1/12 , A3 = 1/18 , A4 = 1/24 4 A1
= 1/4, A2 =1/8 , A3 = 1/12 , A4 = 1/16 Go to Discussion
Solution Qus : 5 2 The value of $\int \frac{({x}^2-1)}{{x}^3\sqrt[]{2{x}^4-2{x}^2+1}}dx$
1 $$2\sqrt[]{2-\frac{2}{{x}^2}+\frac{1}{{x}^4}}+C$$ 2 $$2\sqrt[]{2+\frac{2}{{x}^2}+\frac{1}{{x}^4}}+C$$ 3 $$\frac{1}{2}\sqrt[]{2-\frac{2}{{x}^2}+\frac{1}{{x}^4}}+C$$ 4 None of the above Go to Discussion
Solution Qus : 6 2 The value of $\int \sqrt{x} e^{\sqrt{x}} dx$ is equal to:
1 $$2\sqrt{x}-e^{\sqrt{x}-4\sqrt{xe^{\sqrt{x}}}}+C$$ 2 $$(2x-4\sqrt{x}+4)e^{\sqrt{x}}+C$$ 3 $$(2x+4\sqrt{x}+4)e^{\sqrt{x}}+C$$ 4 $$(1-4\sqrt{x})e^{\sqrt{x}}$$ Go to Discussion
Solution Qus : 7 4 $\int {3}^{{3}^{{3}^x}}.{3}^{{3}^x}.{3}^xdx$ is equal to
1 $\frac{3^{{3}^x}.3^x}{(\log 3){}^3}^{}+c$ 2 $\frac{{3}^3}{(\log 3){}^3}^{}+c$ 3 $\frac{3^{{3}^x}}{(\log 3){}^3}^{}+c$ 4 $\frac{3^{{3}^{{3}^{{}^x}}}}{(\log 3){}^3}^{}+c$ Go to Discussion
Solution Qus : 8 4 The value of $\int \frac{(x+1)}{x(xe^{x}+1)} dx$ is equal to
1 $$log(\frac{1+xe^{x}}{xe^{x}})+C$$ 2 $$log[xe^{x}(1+xe^{x})]+C$$ 3 $$log(\frac{1}{1+xe^{x}})+C$$ 4 $$log(\frac{xe^{x}}{1+xe^{x}})+C$$ Go to Discussion
Solution Qus : 9 3 If $\int e^{x}(f(x)-f'(x))dx=\phi(x)$ , then the value of $\int e^x f(x) dx$ is
1 $$\phi(x)+e^xf(x)$$ 2 $$\phi(x)-e^xf(x)$$ 3 $$\frac{1}{2}[e^xf(x)+\phi(x)]$$ 4 $$\frac{1}{2}[e^xf'(x)+\phi(x)]$$ Go to Discussion
Solution Qus : 11 3 Evaluate
1 ex cosx + C 2 ex secx tanx + C 3 ex tanx + C 4 ex cosx - 1 + C Go to Discussion
Solution Qus : 12 2 If $ \int \frac{xe^{x}}{\sqrt{1+e^{x}}}=f(x)\sqrt{1+e^{x}}-2log \frac{\sqrt{1+e^{x}}-1}{\sqrt{1+e^{x}}+1}+C$ then $f(x)$ is
1 $2x-1$ 2 $2x-4$ 3 $x+4$ 4 $x-1$ Go to Discussion
Solution [{"qus_id":"3916","year":"2019"},{"qus_id":"3906","year":"2019"},{"qus_id":"9453","year":"2020"},{"qus_id":"9454","year":"2020"},{"qus_id":"10709","year":"2021"},{"qus_id":"11148","year":"2022"},{"qus_id":"11522","year":"2023"},{"qus_id":"11539","year":"2023"},{"qus_id":"10214","year":"2015"},{"qus_id":"10442","year":"2014"},{"qus_id":"10457","year":"2014"},{"qus_id":"10485","year":"2014"}]
, then the values of A1, A2, A3, A4 are
