CUET UG Mathematics Applied Indefinite Integration Previous Year Questions (PYQs)
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$\int \frac{dx}{x(x^5+3)}$ is equal to
1
2
3
4
Solution
Let
$I=\int \frac{dx}{x(x^5+3)}$
Put $t=x^5$
Then
$dt=5x^4 dx$
Rewrite integral using substitution:
$I=\frac{1}{5}\int \frac{dt}{t(t+3)}$
Now use partial fractions:
$\frac{1}{t(t+3)}=\frac{1}{3}\left(\frac{1}{t}-\frac{1}{t+3}\right)$
So,
$I=\frac{1}{5}\cdot\frac{1}{3}\int\left(\frac{1}{t}-\frac{1}{t+3}\right)dt$
$I=\frac{1}{15}\log\left|\frac{t}{t+3}\right|+C$
Substitute back $t=x^5$
$I=\frac{1}{15}\log\left|\frac{x^5}{x^5+3}\right|+C$
If $\int \frac{x^3}{x+1},dx=q(x)-\log|x+1|+C$ then $q(x)$ is equal to:
1
2
3
4
Solution
Divide:
$\frac{x^3}{x+1}=x^2-x+1-\frac{1}{x+1}$
So,
$\int \frac{x^3}{x+1},dx=\int (x^2-x+1),dx-\int \frac{1}{x+1},dx$
$=\frac{x^3}{3}-\frac{x^2}{2}+x-\log|x+1|+C$
Comparing with
$q(x)-\log|x+1|+C$
We get
$q(x)=\frac{x^3}{3}-\frac{x^2}{2}+x$
Match List I with List II
| List I | List II |
| A. $\int \frac{1}{x+\sqrt{x}}\,dx$ | I. $2\sqrt{x}+C$ |
| B. $\int \frac{e^{\log\sqrt{x}}}{x}\,dx$ | II. $2(\sqrt{x}-1)e^{\sqrt{x}}+C$ |
| C. $\int \frac{dx}{4x^2-9}$ | III. $2\log(\sqrt{x}+1)+C$ |
| D. $\int e^{\sqrt{x}}\,dx$ | IV. $\frac{1}{12}\log\left(\frac{2x-3}{2x+3}\right)+C$ |
Choose the correct answer from the options given below:
1
2
3
4
Solution
A. $\int \frac{1}{x+\sqrt{x}}dx$
Let $t=\sqrt{x}$
$x=t^2,\quad dx=2t\,dt$
Integral becomes
$\int \frac{2t}{t^2+t}dt
=2\int \frac{1}{t+1}dt
=2\log(t+1)+C
=2\log(\sqrt{x}+1)+C$
So
A → III
B. $\int \frac{e^{\log\sqrt{x}}}{x}dx$
Since $e^{\log\sqrt{x}}=\sqrt{x}$
$\int \frac{\sqrt{x}}{x}dx
=\int x^{-1/2}dx
=2\sqrt{x}+C$
So
B → I
C. $\int \frac{dx}{4x^2-9}$
Using standard formula
$\int \frac{dx}{a^2x^2-b^2}
=\frac{1}{2ab}\log\left|\frac{ax-b}{ax+b}\right|+C$
Here $a=2,\ b=3$
Result
$\frac{1}{12}\log\left(\frac{2x-3}{2x+3}\right)+C$
So
C → IV
D. $\int e^{\sqrt{x}}dx$
Let $t=\sqrt{x}$
$x=t^2,\ dx=2t\,dt$
Integral
$\int e^{\sqrt{x}}dx
=2\int te^t dt$
Using integration by parts
$2(t-1)e^t+C$
Substitute $t=\sqrt{x}$
$2(\sqrt{x}-1)e^{\sqrt{x}}+C$
So
D → II
CUET UG Mathematics Applied
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CUET UG Mathematics Applied
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and More.
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