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Correct ANSWER is (b) \(\FRAC{1}{2} TAN^{-1}⁡\frac{x}{2}+C\)

The EXPLANATION: \(\int \frac{dx}{x^2+4}=\int \frac{dx}{x^2+2^2}\)

Using the formula \(\int \frac{dx}{a^2+x^2}=\frac{1}{a} tan^{-1}⁡\frac{x}{a}+C\)

∴\(\int \frac{dx}{x^2+2^2}=(\frac{1}{2} tan^{-1}⁡\frac{x}{2})+C\)

The correct OPTION is (d) \(\frac{7X^3}{3}-\frac{x^4}{4}+x^2+C\)

EXPLANATION: To find \(\int 7x^2-x^3+2x DX\)

\(\int 7x^2-x^3+2x dx=\int 7x^2 dx-\int x^3 dx+2\int x dx\)

Using \(\int x^n dx=\frac{x^{n+1}}{n+1}\), we get

\(\int 7x^2-x^3+2x dx=\frac{7x^{2+1}}{2+1}-\frac{x^{3+1}}{3+1}+2(\frac{x^{1+1}}{1+1})\)

∴\(\int 7x^2-x^3+2x dx=\frac{7x^3}{3}-\frac{x^4}{4}+x^2+C\)

Correct choice is (b) \(\frac{π}{4}\)

The best I can EXPLAIN: Let \(I=\int_1^{√3} \frac{3}{1+x^2}\)

F(x)=\(\INT \frac{3}{1+x^2}dx\)

=3\(\int \frac{1}{1+x^2} \,dx\)

=3 tan^-1⁡x

Applying the LIMITS, we get

I=F(\(\SQRT{3}\))-F(1)

=3 tan^-1⁡\(\sqrt{3}\)-3 tan^-1⁡1

\(3(\frac{π}{3})-\frac{3π}{4}=\frac{4π-3π}{4}=\frac{π}{4}\).

Right ANSWER is (d) \(\frac{185}{4}\)

The explanation is: Let \(I=\int_3^45x^3 \,dx\)

F(X)=∫ 5x^3 dx

=\(\frac{5x^4}{4}\)

Applying the limits by using the fundamental THEOREM of calculus, we get

I=F(4)-F(3)

=\(\frac{5(4)^3}{4}-\frac{5(3)^3}{4}=\frac{5}{4}(4^3-3^3)\)

=\(\frac{5}{4} (64-27)=\frac{5}{4} (37)=\frac{185}{4}\)

Right option is (c) \(\frac{3π-2}{3}\)

The explanation is: LET \(I=\int_0^π(1-sin⁡3x)DX\)

F(x)=∫ 1-sin⁡3x dx

=x+\(\frac{cos⁡3x}{3}\)

APPLYING the limits by USING the fundamental THEOREM of calculus, we get

I=F(π)-F(0)

=\(π+\frac{cos⁡3π}{3}-0-\frac{cos⁡0}{3}\)

=\(π-\frac{1}{3}-\frac{1}{3}=π-\frac{2}{3}=\frac{3π-2}{3}\).

The correct option is (b) Upper limit

For explanation I would say: In \(\int_a^b\)F(y) dy ‘a’ is the CALLED as lower limit and ‘b’ is called the upper limit of the integral. The FUCTION f in \(\int_a^b\)f(y) dy is called the integrand. The letter ‘y’ is a dummy symbol and can be REPLACED by any other symbol.

Right choice is (b) -cot⁡xe^-x+C

To explain: \(\INT \frac{E^{-x} (1-x)}{sin^2⁡(xe^{-x})} dx\)

LET xe^-x=t

Differentiating w.r.t x, we get

\(-xe^{-x}+e^{-x} dx=DT\)

e^-x (1-x)dx=dt

\(\int \frac{e^{-x} (1-x)}{sin^2⁡(xe^{-x})} dx=\int \frac{dt}{sin^2⁡t}\)

=\(\int cosec^2 \,t \,dt\)

=-cot⁡t+C

Replacing t with xe^-x, we get

\(\int \frac{e^{-x} (1-x)}{sin^2⁡(xe^{-x})} dx=-cot⁡xe^{-x}+C\).

Right choice is (a) \(3 \,sin⁡X-\frac{1}{x}+C\)

The BEST I can EXPLAIN: To find \(\int \,3 \,cos⁡x+\frac{1}{x^2} dx\)

\(\int \,3 \,cos⁡x+\frac{1}{x^2} dx=3 \int cos⁡x \,dx+\int \frac{1}{x^2} \,dx\)

\(\int \,3 \,cos⁡x+\frac{1}{x^2} dx=3 \,sin⁡x+\int x^{-2} \,dx\)

\(\int \,3 \,cos⁡x+\frac{1}{x^2} dx=3 \,sin⁡x+\frac{x^{-2+1}}{-2+1}\)

\(\int \,3 \,cos⁡x+\frac{1}{x^2} dx=3 \,sin⁡x-\frac{1}{x}+C\)

Right choice is (a) \(\FRAC{a^2 \,x^5}{5}+B^2 \,x+\frac{2abx^3}{3}+C\)

EXPLANATION: To find (ax^2+b)^2

\(\int (ax^2+b)^2 dx=\int (a^2 \,x^4+b^2+2ax^2 \,b) dx\)

\(\int (ax^2+b)^2 dx=\int \,a^2 \,x^4 \,dx+\int \,b^2 \,dx+2\int \,ax^2 \,b \,dx\)

\(\int (ax^2+b)^2 dx=a^2 \,\int \,x^4 \,dx+b^2 \int \,dx+2ab\int \,x^2 \,dx\)

\(\int (ax^2+b)^2 dx=a^2 (\frac{x^5}{5})+b^2 x+2ab(\frac{x^3}{3})\)

\(\int (ax^2+b)^2 dx=\frac{a^2 \,x^5}{5}+b^2 x+\frac{2abx^3}{3}+C\)

Right ANSWER is (d) 5-\(\frac{5}{\sqrt{2}}\)

Easy explanation: I=\(\int_0^1 \frac{5 \,sin⁡(TAN^{-1)}x}{1+x^2} \,dx\)

LET tan^-1⁡x=t

Differentiating w.r.t x, we get

\(\frac{1}{1+x^2} \,dx=dt\)

The new limits

When x=0, t=tan^-1⁡0=0

When x=1, t=tan^-1)1=π/4

∴\(\int_0^1 \frac{5 \,sin⁡(tan^{-1}⁡x)}{1+x^2} \,dx=\int_0^{π/4} \,5 \,sin⁡t \,dt\)

=\(5[-cos⁡t]_0^{π/4}=-5[cos⁡t]_0^{π/4}\)

\(=-5(cos⁡ \frac{π}{4}-cos⁡0)=-5(\frac{1}{\sqrt{2}-1})=5-\frac{5}{\sqrt{2}}\)

Right choice is (a) Reverse integral PROPERTY

The BEST explanation: \(\int^1_{-1}\)sin⁡X dx=-\(\int_1^{-1}\)sin⁡x dx comes under the reverse integral property.

In the reverse integral property the upper limits and lower limits are INTERCHANGED. The reverse integral property of definite integrals is \(\int_a^b\)f(x)dx=-\(\int_b^a\)f(x)dx.

Right ANSWER is (d) \(\frac{e^4-1}{2}\)

To ELABORATE: LET \(I=\int_0^2 \,e^2x \,dx\)

F(X)=\(\int e^{2x} dx\)

=\(\frac{e^{2x}}{2}\)

Applying the limits, we get

I=F(2)-F(0)

=\(\frac{e^2(2)}{2}-\frac{e^2(0)}{2}=\frac{(e^4-1)}{2}\).

Right CHOICE is (d) \(\FRAC{sin⁡2x}{2}\)-X cos⁡2x+C

To explain: By using ∫ u.v DX=u∫ v dx-∫ u'(∫ v dx)

∫ x TANX dx=2x ∫ sin⁡2x dx-∫ (2x)’ ∫ sin⁡2x dx

=2x (-\(\frac{cos⁡2x}{2}\))+2\(\int \frac{cos⁡2x}{2}\) dx

=-xcos 2x+∫ cos⁡2x dx

=\(\frac{sin⁡2x}{2}\)-x cos⁡2x+C

The CORRECT CHOICE is (B) 5(x+cos⁡x)+C

For explanation I would say: \(\int \frac{5 cos^2⁡x}{1+sin⁡x} DX=\int \frac{5(1-sin^2⁡x)}{1+sin⁡x}=5\int \frac{(1+sin⁡x)(1-sin⁡x)}{(1+sin⁡x)} dx\)

=5∫ (1-sin⁡x)dx

=5(x-(-cos⁡x))=5(x+cos⁡x)+C

The CORRECT option is (b) –\(\frac{cos⁡2x}{2}+\frac{e^{3x}}{3}-\frac{sin⁡3x}{3}+C\)

To explain I would SAY: To find \(\INT \,sin⁡2x+e^{3x}-cos⁡3x \,DX\)

\(\int sin⁡2x+e^{3x}-cos⁡3x \,dx=\int \,sin⁡2x \,dx+\int \,e^{3x} \,dx-\int \,cos⁡3x \,dx\)

\(\int sin⁡2x+e^{3x}-cos⁡3x \,dx=-\frac{cos⁡2x}{2}+\frac{e^{3x}}{3}-\frac{sin⁡3x}{3}+C\)

Right CHOICE is (c) -cos⁡2x+3x+C

Best EXPLANATION: To find ∫ 2 sin⁡2x+3 DX

\(\int \,2 \,sin⁡2x+3 \,dx=\int \,2 \,sin⁡2x \,dx + \int \,3 \,dx\)

\(\int \,2 \,sin⁡2x+3 \,dx=2\int \,sin⁡2x \,dx+3\int \,dx\)

\(\int \,2 \,sin⁡2x+3 \,dx=\frac{-2 cos⁡2x}{2}+3x\)

∴∫2 sin⁡2x+3 dx=-cos⁡2x+3x+C

The CORRECT CHOICE is (b) 2x^2-3 log⁡x+C

Easiest EXPLANATION: To find \(\int \frac{4x^4-3x^2}{x^3} dx\)

\(\int \frac{4x^4-3x^2}{x^3} \,dx=\int \frac{4x^4}{x^3} – \frac{3x^2}{x^3} \,dx\)

\(\int \frac{4x^4-3x^2}{x^3} \,dx=\int 4x dx-\int \frac{3}{x} dx\)

\(\int \frac{4x^4-3x^2}{x^3} \,dx=\frac{4x^2}{2}-3 log⁡x\)

∴ \(\int \frac{4x^4-3x^2}{x^3} \,dx=2x^2-3 \,log⁡x+C\).