(A) : \(I=\int \frac{x^{2}+1}{x^{4}-x^{2}+1} d x\) \(=\int \frac{1+\frac{1}{x^{2}}}{x^{2}-1+\frac{1}{x^{2}}} d x=\int \frac{1+\frac{1}{x^{2}}}{\left(x-\frac{1}{x}\right)^{2}+1} d x\) Let, \(\quad \mathrm{x}-\frac{1}{\mathrm{x}}=\mathrm{t}\) \(\left(1+\frac{1}{x^{2}}\right) d x=d t\) \(\therefore \quad \mathrm{I}=\int \frac{\mathrm{dt}}{\mathrm{t}^{2}+1}=\tan ^{-1}(\mathrm{t})+\mathrm{c}\) \(=\tan ^{-1}\left(\mathrm{x}-\frac{1}{\mathrm{x}}\right)+\mathrm{c}=\tan ^{-1}\left(\frac{\mathrm{x}^{2}-1}{\mathrm{x}}\right)+\mathrm{c}\)
MHT CET-2019
Integral Calculus
86244
\(\int_{a}^{b} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{a+b-x}} d x=\)
1 \(\mathrm{a}-\mathrm{b}\)
2 \(\frac{b-a}{2}\)
3 \(\frac{a-b}{2}\)
4 \(a+b\)
Explanation:
(B) : \(I=\int_{a}^{b} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{a+b-x}} d x\) \(I=\int_{a}^{b} \frac{\sqrt{a+b-x}}{\sqrt{a+b-x}+\sqrt{a+b-(a+b-x}} d x I=\int_{a}^{b} \frac{\sqrt{a+b-x}}{\sqrt{a+b-x}+\sqrt{x}} d x \tag{ii}\) From equation (i) and equation (ii) adding, we get - \(2 I=\int_{a}^{b} \frac{\sqrt{x}+\sqrt{a+b-x}}{\sqrt{x}+\sqrt{a+b-x}} d x\) \(2 I=\int_{a}^{b} d x=[x]_{a}^{b}=b-a\) \(I=\frac{b-a}{2}\)
MHT CET-2019
Integral Calculus
86245
If \(\int \frac{d x}{\sqrt{16-9 x^{2}}}=A \sin ^{-1}(B x)+C\) then \(A+B=\)
1 \(\frac{9}{4}\)
2 \(\frac{19}{4}\)
3 \(\frac{3}{4}\)
4 \(\frac{13}{12}\)
Explanation:
(D) : Given,\(\int \frac{\mathrm{dx}}{\sqrt{16-9 \mathrm{x}^{2}}}=\mathrm{A} \sin ^{-1}(\mathrm{Bx})+\mathrm{C}\) L.H.S., \(\mathrm{I}=\int \frac{\mathrm{dx}}{\sqrt{16-9 \mathrm{x}^{2}}}\) \(=\frac{1}{3} \int \frac{\mathrm{dx}}{\sqrt{\left(\frac{4}{3}\right)^{2}-\mathrm{x}^{2}}}=\frac{1}{3} \sin ^{-1} \frac{3 \mathrm{x}}{4}+\mathrm{C}\) On comparing L.H.S. and R.H.S., we get- \(\mathrm{A}=\frac{1}{3}, \mathrm{~B}=\frac{3}{4}\) \(\therefore \quad \mathrm{A}+\mathrm{B}=\frac{1}{3}+\frac{3}{4}=\frac{13}{12}\)
MHT CET-2018
Integral Calculus
86246
\(\int \mathrm{e}^{x}\left[\frac{2+\sin 2 x}{1+\cos 2 x}\right] d x=\)
(A) : \(I=\int \frac{x^{2}+1}{x^{4}-x^{2}+1} d x\) \(=\int \frac{1+\frac{1}{x^{2}}}{x^{2}-1+\frac{1}{x^{2}}} d x=\int \frac{1+\frac{1}{x^{2}}}{\left(x-\frac{1}{x}\right)^{2}+1} d x\) Let, \(\quad \mathrm{x}-\frac{1}{\mathrm{x}}=\mathrm{t}\) \(\left(1+\frac{1}{x^{2}}\right) d x=d t\) \(\therefore \quad \mathrm{I}=\int \frac{\mathrm{dt}}{\mathrm{t}^{2}+1}=\tan ^{-1}(\mathrm{t})+\mathrm{c}\) \(=\tan ^{-1}\left(\mathrm{x}-\frac{1}{\mathrm{x}}\right)+\mathrm{c}=\tan ^{-1}\left(\frac{\mathrm{x}^{2}-1}{\mathrm{x}}\right)+\mathrm{c}\)
MHT CET-2019
Integral Calculus
86244
\(\int_{a}^{b} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{a+b-x}} d x=\)
1 \(\mathrm{a}-\mathrm{b}\)
2 \(\frac{b-a}{2}\)
3 \(\frac{a-b}{2}\)
4 \(a+b\)
Explanation:
(B) : \(I=\int_{a}^{b} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{a+b-x}} d x\) \(I=\int_{a}^{b} \frac{\sqrt{a+b-x}}{\sqrt{a+b-x}+\sqrt{a+b-(a+b-x}} d x I=\int_{a}^{b} \frac{\sqrt{a+b-x}}{\sqrt{a+b-x}+\sqrt{x}} d x \tag{ii}\) From equation (i) and equation (ii) adding, we get - \(2 I=\int_{a}^{b} \frac{\sqrt{x}+\sqrt{a+b-x}}{\sqrt{x}+\sqrt{a+b-x}} d x\) \(2 I=\int_{a}^{b} d x=[x]_{a}^{b}=b-a\) \(I=\frac{b-a}{2}\)
MHT CET-2019
Integral Calculus
86245
If \(\int \frac{d x}{\sqrt{16-9 x^{2}}}=A \sin ^{-1}(B x)+C\) then \(A+B=\)
1 \(\frac{9}{4}\)
2 \(\frac{19}{4}\)
3 \(\frac{3}{4}\)
4 \(\frac{13}{12}\)
Explanation:
(D) : Given,\(\int \frac{\mathrm{dx}}{\sqrt{16-9 \mathrm{x}^{2}}}=\mathrm{A} \sin ^{-1}(\mathrm{Bx})+\mathrm{C}\) L.H.S., \(\mathrm{I}=\int \frac{\mathrm{dx}}{\sqrt{16-9 \mathrm{x}^{2}}}\) \(=\frac{1}{3} \int \frac{\mathrm{dx}}{\sqrt{\left(\frac{4}{3}\right)^{2}-\mathrm{x}^{2}}}=\frac{1}{3} \sin ^{-1} \frac{3 \mathrm{x}}{4}+\mathrm{C}\) On comparing L.H.S. and R.H.S., we get- \(\mathrm{A}=\frac{1}{3}, \mathrm{~B}=\frac{3}{4}\) \(\therefore \quad \mathrm{A}+\mathrm{B}=\frac{1}{3}+\frac{3}{4}=\frac{13}{12}\)
MHT CET-2018
Integral Calculus
86246
\(\int \mathrm{e}^{x}\left[\frac{2+\sin 2 x}{1+\cos 2 x}\right] d x=\)
(A) : \(I=\int \frac{x^{2}+1}{x^{4}-x^{2}+1} d x\) \(=\int \frac{1+\frac{1}{x^{2}}}{x^{2}-1+\frac{1}{x^{2}}} d x=\int \frac{1+\frac{1}{x^{2}}}{\left(x-\frac{1}{x}\right)^{2}+1} d x\) Let, \(\quad \mathrm{x}-\frac{1}{\mathrm{x}}=\mathrm{t}\) \(\left(1+\frac{1}{x^{2}}\right) d x=d t\) \(\therefore \quad \mathrm{I}=\int \frac{\mathrm{dt}}{\mathrm{t}^{2}+1}=\tan ^{-1}(\mathrm{t})+\mathrm{c}\) \(=\tan ^{-1}\left(\mathrm{x}-\frac{1}{\mathrm{x}}\right)+\mathrm{c}=\tan ^{-1}\left(\frac{\mathrm{x}^{2}-1}{\mathrm{x}}\right)+\mathrm{c}\)
MHT CET-2019
Integral Calculus
86244
\(\int_{a}^{b} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{a+b-x}} d x=\)
1 \(\mathrm{a}-\mathrm{b}\)
2 \(\frac{b-a}{2}\)
3 \(\frac{a-b}{2}\)
4 \(a+b\)
Explanation:
(B) : \(I=\int_{a}^{b} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{a+b-x}} d x\) \(I=\int_{a}^{b} \frac{\sqrt{a+b-x}}{\sqrt{a+b-x}+\sqrt{a+b-(a+b-x}} d x I=\int_{a}^{b} \frac{\sqrt{a+b-x}}{\sqrt{a+b-x}+\sqrt{x}} d x \tag{ii}\) From equation (i) and equation (ii) adding, we get - \(2 I=\int_{a}^{b} \frac{\sqrt{x}+\sqrt{a+b-x}}{\sqrt{x}+\sqrt{a+b-x}} d x\) \(2 I=\int_{a}^{b} d x=[x]_{a}^{b}=b-a\) \(I=\frac{b-a}{2}\)
MHT CET-2019
Integral Calculus
86245
If \(\int \frac{d x}{\sqrt{16-9 x^{2}}}=A \sin ^{-1}(B x)+C\) then \(A+B=\)
1 \(\frac{9}{4}\)
2 \(\frac{19}{4}\)
3 \(\frac{3}{4}\)
4 \(\frac{13}{12}\)
Explanation:
(D) : Given,\(\int \frac{\mathrm{dx}}{\sqrt{16-9 \mathrm{x}^{2}}}=\mathrm{A} \sin ^{-1}(\mathrm{Bx})+\mathrm{C}\) L.H.S., \(\mathrm{I}=\int \frac{\mathrm{dx}}{\sqrt{16-9 \mathrm{x}^{2}}}\) \(=\frac{1}{3} \int \frac{\mathrm{dx}}{\sqrt{\left(\frac{4}{3}\right)^{2}-\mathrm{x}^{2}}}=\frac{1}{3} \sin ^{-1} \frac{3 \mathrm{x}}{4}+\mathrm{C}\) On comparing L.H.S. and R.H.S., we get- \(\mathrm{A}=\frac{1}{3}, \mathrm{~B}=\frac{3}{4}\) \(\therefore \quad \mathrm{A}+\mathrm{B}=\frac{1}{3}+\frac{3}{4}=\frac{13}{12}\)
MHT CET-2018
Integral Calculus
86246
\(\int \mathrm{e}^{x}\left[\frac{2+\sin 2 x}{1+\cos 2 x}\right] d x=\)
(A) : \(I=\int \frac{x^{2}+1}{x^{4}-x^{2}+1} d x\) \(=\int \frac{1+\frac{1}{x^{2}}}{x^{2}-1+\frac{1}{x^{2}}} d x=\int \frac{1+\frac{1}{x^{2}}}{\left(x-\frac{1}{x}\right)^{2}+1} d x\) Let, \(\quad \mathrm{x}-\frac{1}{\mathrm{x}}=\mathrm{t}\) \(\left(1+\frac{1}{x^{2}}\right) d x=d t\) \(\therefore \quad \mathrm{I}=\int \frac{\mathrm{dt}}{\mathrm{t}^{2}+1}=\tan ^{-1}(\mathrm{t})+\mathrm{c}\) \(=\tan ^{-1}\left(\mathrm{x}-\frac{1}{\mathrm{x}}\right)+\mathrm{c}=\tan ^{-1}\left(\frac{\mathrm{x}^{2}-1}{\mathrm{x}}\right)+\mathrm{c}\)
MHT CET-2019
Integral Calculus
86244
\(\int_{a}^{b} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{a+b-x}} d x=\)
1 \(\mathrm{a}-\mathrm{b}\)
2 \(\frac{b-a}{2}\)
3 \(\frac{a-b}{2}\)
4 \(a+b\)
Explanation:
(B) : \(I=\int_{a}^{b} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{a+b-x}} d x\) \(I=\int_{a}^{b} \frac{\sqrt{a+b-x}}{\sqrt{a+b-x}+\sqrt{a+b-(a+b-x}} d x I=\int_{a}^{b} \frac{\sqrt{a+b-x}}{\sqrt{a+b-x}+\sqrt{x}} d x \tag{ii}\) From equation (i) and equation (ii) adding, we get - \(2 I=\int_{a}^{b} \frac{\sqrt{x}+\sqrt{a+b-x}}{\sqrt{x}+\sqrt{a+b-x}} d x\) \(2 I=\int_{a}^{b} d x=[x]_{a}^{b}=b-a\) \(I=\frac{b-a}{2}\)
MHT CET-2019
Integral Calculus
86245
If \(\int \frac{d x}{\sqrt{16-9 x^{2}}}=A \sin ^{-1}(B x)+C\) then \(A+B=\)
1 \(\frac{9}{4}\)
2 \(\frac{19}{4}\)
3 \(\frac{3}{4}\)
4 \(\frac{13}{12}\)
Explanation:
(D) : Given,\(\int \frac{\mathrm{dx}}{\sqrt{16-9 \mathrm{x}^{2}}}=\mathrm{A} \sin ^{-1}(\mathrm{Bx})+\mathrm{C}\) L.H.S., \(\mathrm{I}=\int \frac{\mathrm{dx}}{\sqrt{16-9 \mathrm{x}^{2}}}\) \(=\frac{1}{3} \int \frac{\mathrm{dx}}{\sqrt{\left(\frac{4}{3}\right)^{2}-\mathrm{x}^{2}}}=\frac{1}{3} \sin ^{-1} \frac{3 \mathrm{x}}{4}+\mathrm{C}\) On comparing L.H.S. and R.H.S., we get- \(\mathrm{A}=\frac{1}{3}, \mathrm{~B}=\frac{3}{4}\) \(\therefore \quad \mathrm{A}+\mathrm{B}=\frac{1}{3}+\frac{3}{4}=\frac{13}{12}\)
MHT CET-2018
Integral Calculus
86246
\(\int \mathrm{e}^{x}\left[\frac{2+\sin 2 x}{1+\cos 2 x}\right] d x=\)
